LIBRARY OF CONGRESS. 



\ K2.„S-5 



Chap. Copyright No.. 

Shelfj"^^ 



UNITED STATES OF AMERICA. 



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1 



THE ACTION OF MATERIALS UNDER STRESS 



OR 



STRUCTURAL MECHANICS 



COMPRISING THE 



STRENGTH AND RESISTANCE OF MATERIALS AND 
ELEMENTS OF STRUCTURAL DESIGN 



Wrni EXAMPLES AND PROBLEMS 



BY 

CHARLES E. GREENE, A.NL, C.E., 

PKOFESCOR OF CIVIL F,NGINE£RING, UNIV'KKblTY OF MICHIGAN 
CONSULTING ENGINEER. 



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ANN ARBOR, MICH.: 

Printed for the Author 
1S97. 



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BY THE SAME AUTHOR. 

Graphics for Engineers, Architects and Builders, 
A manual for designers, and a text-book for scien- 
tific schools. 

Trusses and Arches; Analyzed and Discussed by 
Graphical Methods. In three parts — published 
separately. 

Part I. Roof Trusses: Diagrams for Steady Load, 
Snow aud Wind. 8vo., 80 pp., 3 folding plates. 
Revised Edition. $1.25. 

Part II. Bridge Trusses: Single, Continuous, and 
Draw Spans; Single and Multiple Systems; Straight 
and Inclined Chords. 8 vo., igo pp., lO folding 
plates. Fifth Edition Revised. $2.50. 

Part III. Arches; in Wood, Iron, and Stone, for 
Roofs, Bridges, and Wall-Openings; Arched Ribs, 
and Braced Arches; Stresses from Wind and Change 
of Temperature. 8 vo., 160 pp., 8 folding plates. 
Second Edition Revised. $2.50. 



Copyright, 1897, 
By Charles E. Greene. 




THE INLAND PRESS, ANN ARBOR. 




^ 



PREFACE. 

The author, in teaching for many years the subjects 
embraced in the following pages, has found it advantageous 
to take at first but a portion of what is included in the sev- 
eral chapters, and, after a general survey of the field, to return 
and extend the investigation more in detail. Some of the 
sections, therefore, are printed in smaller type and can be 
omitted at first reading. A few of the special investigations 
may become of interest only when the problems to which they 
relate occur in actual practice. 

It is hoped that this book will be serviceable after the 
class-room work is concluded, and reference is facilitated by 
a more compact arrangement of the several matters than the 
course suggested above would give. The attempt has been 
made to deal with practicable cases, and the examples for the 
most part are shaped with that end in view. A full index will 
enable one to find any desired topic. 

The treatment of the subject of internal stress is largely 
graphical. All the constructions are simple, and the results, 
besides being useful in themselves, shed much light on various 
problems. The time devoted to a careful study of the chap- 
ter in question will be well expended. 

The notation is practically uniform throughout the book, 
and is that used by several standard authors. Forces and 
moments are expressed by capital letters, and unit loads and 
stresses by small letters. The co-ordinate x is measured along 
the length of a piece, the co-ordinate y in the direction of 
variation of stress in a section, and z is the line of no varia- 
tion of stress, that is, the line parallel to the moment axis. 

One who has mastered the subjects discussed here can 
use the current formulas, the pocket-book rules, and tables, 
not blindly, but with discrimination, and ought to be prepared 
to design intelligently. 

Mr. Albert E. Greene has rendered much assistance in 
preparing the material for publication. 



TABLE OF CONTENTS. 



Introduction. 


Chapter 


L 


Chapter 


IL 


Chapter 


in. 


Chapter 


IV. 


Chapter 


V. 


Chapter 


VI. 


Chapter 


VII. 


Chapter 


VIII. 


Chapter 


IX. 


Chapter 


X. 


Chapter 


XI. 


Chapter 


XII. 


Chapter XIIL 


Chapter 


XIV 


Chapter 


XV 


Chapter XVL 



PAGd 
I 



Action of a Piece under Direct Force, 6 

Materials, - - - - - 19 

Beams, - - - - - - 44 

Torsion, - - - - - - 80 

Moments of Inertia, - - - - 85 

Flexure and Deflection of Simple Beams 96 

Restrained Beams: Continuous Beams, 113 

Pieces under Tension, - - - 129 
Compression Pieces: — Columns, Posts 

and Struts, - - - - - 138 

Safe Working Stresses, - - - I57 

Internal Stress: Change of Form, - 183 

Rivets: Pins, ----- 206 

Envelopes: — Boilers, Pipes, Dome, - 217 

Plate Girder, _ _ _ - _ 240 
Earth Pressure: Retaining Wall: Springs: 

Plates, ------ 247 

Details in Wood and Iron, - - 258 



INTRODUCTION. 

I. External Forces. — The engineer, in designing a new 
structure, or critically examining one already built, determines 
from the conditions of the case the actual or probable external 
forces which the structure is called upon to resist. He may 
then prepare, either b}^ mathematical calculations or by 
graphical methods,^ a sheet which shows the maximum and 
minimum direct forces of tension and compression which the 
several pieces or parts of the structure are liable to experience,- 
as well as the bending moments on such parts as are sub- 
jected to them. 

These forces and moments are determined from the re- 
quirements of equilibrium, if the pieces are at rest. For forces 
acting in one plane, a condition which suffices for the analysis 
of most cases, it is necessary that, for the structure as a 
whole, as well as for each piece, there shall be no tendency to 
move up or down, to move to the right or left, or to rotate. 
These limitations are usually expressed in Mechanics as, that 
the sum of the X forces, the sum of the Y forces, and the sum 
of the moments shall each equal zero. 

If the structure is a machine, the forces and moments in 
action at any time, and their respective magnitudes, call for a 
consideration of the question of acceleration or retardation of 
the several parts and the additional maximum forces and mo- 
ments called into action by the greatest rate of change of 
motion at any instant. Hence the weight or mass of the 
moving part or parts is necessarily taken into account. 

Finally, noting the rapidity and frequency of the change 
of force and moment at any section of any piece or connec- 
tion, the engineer selects, as judgment dictates, the allowable 
stresses of the several kinds per square inch, making allow- 
ance for the effect of impact, shock and vibration in intensify- 

*See the author's "Trusses and Arches," considered Graphically: Part I. 
Roof Trusses; Part II., Bridge Trusses; Part III., Arches. New York, Wiley & 
Sons. 



1 



2 STRUCTURAL MECHANICS. 

ing their action, and proceeds to find the necessary cross-sec- 
tions of the parts and the proportions of the connections 
between them. As all structures are intended to endure the 
forces and vicissitudes to which they are usually exposed, the 
allowable unit-stresses, expressed in pounds per square inch, 
must be safe stresses. 

It is largely with the development of the latter part of 
this subject, after the forces have been found to which the 
several parts are liable, that this book is concerned. 

2. Ties, Struts and Beams.— There are, in general, 
three kinds of pieces in a frame or structure; ties or tension 
members; columns, posts and struts or compression members; 
and beams, which support a transverse load and are subject to 
bending and its accompanying shear. A given piece may also 
be, at the same time, a tie and a beam, or a strut and a beam, 
and at different times a tie and a strut. 

3. Relation of External Forces to Internal Stresses. — 
The forces and moments which a member is called upon to 
resist, and which may properly be considered as external to 
that member, give rise to actions between all the particles of 
material of which such a member is composed, tending to 
move adjacent particles from, towards or by one another, and 
causing change of form. There result internal stresses, or 
resistances to displacement, between the several particles. 

These internal stresses, or briefly stresses, must be of 
such kind, magnitude, distribution and direction, at any 
imaginary section of a piece or structure, that their resultant 
force and moment will satisfy the requirements of equilibrium 
or change of motion with the external resultant force and mo- 
ment at that section; and no stress per square inch can, for a 
correct design, be greater than the material will safely bear. 
Hence may be determined the necessary area and form of the 
cross-section at the critical points, when the resultant forces 
and moments are known. 

4. Internal Stress. — There are three kinds of stress, or 
action of adjacent particles one on the other, to which the 
particles of a body may be subjected, when external forces 
and its own weight are considered, viz. : tensile stress, tending 
to remove one particle farther from its neighbor, and mani- 



INTRODUCTION. 



fested by an accompanying stretch or elongation of the body; 
compressive stress, tending to make a particle approach its 
neighbor, and manifested by an accompanying shortening or 
compression of the body; and shearing stress, tending to make 
a particle move or slide laterally with reference to an adjacent 
particle, and manifested by an accompanying distortion. 
Whether the stress produces change of form, or the attempted 
change of form gives rise to internal stresses as resistances, is 
of little consequence; the stress between two particles and the 
change of position of the particles are always associated, and 
one being given the other must exist. 

5. Tension and Shear, or Compression and Shear. — 
If the direction of the stress is oblique, that is, not normal or 
perpendicular, on any section of a body, the stress may be 
resolved into a tensile or compressive stress normal to that 
section, and a tangential stress along the section, which, 
from its tendency to cause sliding of one portion of the body 
by or along the section, has been given the name of shear, 
from the resemblance to the action of a pair of shears, one 
blade passing by the other along the opposite sides of the 
plane of section. Draw two oblique and directly opposed 
arrows, one on either side of a straight line representing 
the trace of a sectional plane, decompose those oblique 
stresses normally and tangentially to the plane, and notice 
the resulting directly opposed tension or compression, 
and the shear. Hence tension and shear, or compression and 
shear, may be found on any given plane in a body, but tension 
and compression cannot simultaneously occur at one point in 
a given area. 

6. Sign of Stress. — Ties ■ are usually slender members; 
struts have larger lateral dimensions. Longitudinal tension 
tends to diminish the cross-section of the piece which carries 
it, and hence may conveniently be represented by — , the nega- 
tive sign; longitudinal compression tends to increase the cross- 
sectional area and may be called + or positive. Shear, being 
at right angles to the tension and compression in the pre- 
ceding illustration, has no -sign; and lies, in significance, be- 
tween tension and compression. If a rectangular plate is 
pulled in the direction of two -of its opposite sides and com- 



4 STRUCTURAL MECHANICS. 

pressed in the direction of its other two sides, there will be 
some shearing stress on every plane of section except those 
parallel to the sides, and nothing but shear on two certain 
oblique planes, as will be seen later. 

7. Unit Stresses. — These internal stresses are measured 
by units of pounds and inches by English and American en- 
gineers, and are stated as so many pounds of tension, com- 
pression or shear per square inch, called unit tension, com- 
pression or shear. Thus, in a bar of four square inches 
cross-section, under a total pull of 36,000 pounds centrally 
applied, the internal unit tension is 9000 pounds per square 
inch, provided the pull is uniformly distributed on the particles 
adjacent to any cross-section. If the pull is not central or 
the stress not uniformly distributed, the average or mean unit 
tensile stress is still 9000 pounds. 

If an oblique section of the same bar is made, the total 
force acting on the particles adjacent to the section is the 
same as before, but the area of section is increased; hence the 
unit stress, found by dividing the force by the new area, is 
diminished. The stress will also be oblique to the section, as 
its direction rhust be that of the force. When the unit stress 
is not normal to the plane of section on which it acts, it can 
be decomposed into a normal unit tension and a unit shear. 
See § 180. 

When the stress varies in magnitude from point to point, 
its amount on any very small area (the infinitesimal area of 
the Calculus) may be divided by that area, and the quotient 
will be the unit stress, or the amount which would exist on a 
square inch, if a square inch had the same stress all over it as 
the very small area has. 

8. Unit Stresses on Different Planes not to be Treated 
as Forces. — It will be seen, upon inspection of the results of 
analyses which come later, that unit stresses acting on differ- 
ent planes must not be compounded and resolved as if they 
were forces. But the entire stress upon a certain area, found 
by multiplying the unit stress by that area, is a force, and this 
force may be compounded with other forces or resolved, and 
the new force may then be divided by the new area of action, 
and a new unit stress be thus found. 



INTRODUCTION. 5 

Some persons may be assisted in understanding the 
analysis of problems by representing in a sketch, or mentally, 
the unit stresses at different parts of a cross-section by ordin- 
ates which make up, in their assemblage, a volume. This 
volume, whose base is the cross-section, will represent or be 
proportional to the total force on the section. The position 
of the resultant force or forces, i. e., traversing the centre of 
gravity of the volume, the direction and law of distribution of 
the stress are then quite apparent. 



CHAPTER I. 

ACTION OF A PIECE UNDER DIRECT FORCE. 

9. Change of Length under an Applied Force. — Let a 
uniform bar of steel have a moderate amount of tension 
applied to its two ends. It will be found, upon measurement, 
to have increased in length uniformly throughout the measured 
distance. Upon release of the tension the stretch disappears, 
the bar resuming its original length. A second application 
of the same amount of tension will cause the same elongation, 
and its removal will be followed by the same contraction to 
the original length. The bar acts like a spring. This elastic 
elongation (or shortening under compression) is manifested by 
all substances which have definite form and are used in con- 
struction; and it is the cause of such changes of shape as struct- 
ures, commonly considered rigid, experience under changing 
loads. The product of the elongation (or shortening) into the 
mean force that produced it is a measure of the work done in 
causing the change of length. As the energy of a moving 
body can be overcome only by work done, the above product 
becomes of practical interest in structures where moving loads, 
shocks and vibrations play an important part. 

10. Modulus of Elasticity. — If the bar of steel is stretched 
with a greater force, but still a moderate one, it is found by 
careful measurement that the elongation has increased with 
the force; and the relationship may be laid down that the 
elongation per linear inch is directly proportional to the unit 
stress on the cross-section per square inch. 

The ratio of the unit stress to the elongation per unit of 
length is denoted by E, which is termed the modulus of elas- 
ticity of the material, and is based, in English and American 
books, upon the pound and inch as units. If P is the total 
tension in pounds applied to the cross-section S measured in 



A PIECE UNDER DIRECT FORCE. 7 

square inches, U is the elongation in inches, produced by the 
tension, in the previously measured length of / inches, 

P P 
E = — ; /I = per inch. 

Hence, if E has been determined for a given material, the 
stretch of a given bar under a given unit stress is easily found. 

Example. — A bar of 6 sq. in. section stretches 0.09 in., in a 
measured length of 120 in., under a pull of 120,000 lbs. 

T- 120000 X 120 /. 

E = = 26,700,000. 

6x 0,09 

If the stress were compressive, a similar modulus would 
result, which will be shown presently to agree with the one 
just derived. 

If one particle is displaced laterally with regard to its 
neighbor, under the action of a shearing stress, a modulus of 
shearing elasticity will be obtained, denoted by C, the ratio of 
the displacement orjdistorlion to the ujiit shear which accom- tr 
panies it. 

11. Stress-Stretch Diagram. — The elongations caused 
in a certain bar, or the stretch per unit of length, may be 
plotted as abscissas, and the corresponding forces producing 
the stretch, or the unit stresses per square inch, . may be used 
as ordinates, defining a certain curve, as represented in Fig. i. 
This curve can be drawn on paper by the specimen itself, 
when in the testing machine, if the paper is moved in one 
direction to correspond with the movement of the poise on the 
weighing arm, and the pencil is moved at right angles by the 
stretch of the specimen. 

A similar diagram can be made for a compression speci- 
men, and may be drawn in the diagonally opposite quadrant. 
Pull will then be rightly represented as of opposite sign to 
thrust, and extension will be laid off in the opposite direction 
to shortening or compression. 

12. Work of Elongation. — If the different unit stresses 
applied to the bar are laid off on O Y as ordinates and the 
resulting stretches per unit of length on O X as abscissas, the 
portion of the diagram near the origin will be found to be a 



8 STRUCTURAL MECHANICS. 

straight line, more or less oblique, according to the scale by 
which the elongations are platted. The elongation varies 
directly as the unit stress, beginning with zero. Hence the 
mean force is -J P, and the work done in stretching a given 
bar with a given force, if the limit of elastic stretch is not ex- 
ceeded, is 

P PV 

Work = - . A/ = _L1. 

2 2ES 

It may be seen that the work done in stretching the bar 
is represented by the area included between the base line or 
axis, the curve O A and the ordinate at A. It also appears 
that E may be looked upon as the tangent of the angle X O A. 
A material of greater resistance to elongation will give an 
angle greater than X O A and vice versa. 

Example. — A bar 20 ft. = 240 in. long and 3 sq. in. in sec- 
tion is to have a stress applied of 10,000 lbs. per sq. in.; if E = 
28,000,000, the work done on the bar will be 

10,000 . ^0,000 . 240 o^ • 11, 

— ? ^-^ -^ = 1,286 m. lbs., 

2 . 28,000,000 

and the stretch will be 1,286-^5,000 = 0.257 in. 

13. Permanent Set. — While the unit stress may be grad- 
ually increased with corresponding increase of stretch, and 
apparently complete recovery of original length when the bar 
is released, there comes a time when very minute and delicate 
measurements show that the elongation has increased in a 
slightly greater degree than has the stress. The line O A at 
and beyond such a point must therefore be a curve, concave 
to the axis of X. If the piece is now relieved from stress, it 
will be found that the bar has become permanently lengthened. 
The amount of this increase of length after removal of stress 
is called set, or permanent set, and the unit stress for which a 
permanent set can first be detected is known as the elastic 
limit. As the elongation itself" is an exceedingly small quan- 
tity, even when measured in a length of many inches, and the 
permanent set is, in the beginning, a quantity far smaller and 
hence more difficult of determination, the place where the 
straight line O A first begins to curve is naturally hard to 
locate, and the accurate elastic limit is therefore uncertain. 



A PIECE UNDER DIRECT FORCE. 9 

Some contend that O A itself is a curve of extreme flatness. 
The common or commercial elastic limit lies much farther up 
the curve, where the permanent set becomes decidedly not- 
able. 

If, after a certain amount of permanent set has occurred 
in a bar, and the force which caused it has been removed, a 
somewhat smaller force is repeatedly applied to the bar, the 
piece will elongate and contract elastically to the new length, 
i. e., old length plus permanent set, just as if the unit stress 
were below the elastic limit. 



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14. Yield Point. — The unit stress increasing, the elon- 
gation increases and the permanent set increases until a unit 
stress B is reached, known as the yield point (or commercial 
elastic limit, or common elastic limit), which causes the bar 
to yield or draw out without increase of force, and, as the sec- 
tion must decrease, apparently with decreasing power of resist- 
ance. There will then be a break of continuity in the graphic 
curve. A decided permanent elongation of the bar takes 
place at this time — sufficient to dislodge the scale from the 
surface of a steel bar, if left as it comes from the rolls or ham- 
mer. The weighing beam of the testing machine falls, from 
the diminished resistance just referred to, and remains station- 



lO STRUCTURAL MECHANICS. 

ary while the bar is elongating for a sensible interval of time. 
Hence, for steel, the yield point, or common elastic limit, is 
easily determined by what is known as the "drop of the 
beam." The remainder of the curve, up to the breaking 
point, is shown in the figure. 

15. Elastic Limit Raised. — For stresses above the yield 
point also, a second application and release of stress will give 
an elastic elongation and contraction as before the occurrence 
of set, as shown by lines C D, E F, so that a new elastic 
limit may be said to be established. The stretch due to any 
given stress may be considered to be the elastic elongation 
plus the permanent set; and, for repetitions of lesser forces, 
the bar will give a line- parallel to O A, as if drawn from 'a 
new origin on O X, distant from O the amount of the perman- 
ent set. 

The effects of a period of rest and of application of com- 
pression are spoken of later. 

If the line O A is prolonged upwards, it will divide each 
abscissa into two parts, of which that on the left of O A will 
be the elastic stretch, and that on the right of O A the per- 
manent set for a given unit stress. 

16. ^A^ork of Elongation, for Stress above Yield Point. — 
The area below the curve, and limited by any ordinate G D, 
will be the work done in stretching the bar with a force 
represented by the product of that ordinate into the bar's 
cross-section, and if a line be drawn from the upper end of that 
ordinate parallel to O A, the triangle C D G will give the work 
done in elastic stretch and the quasi-parallelogram O B D C 
will show the permanent work of deformation done on the bar. 
It should be remembered that, as the bar stretches, the section 
decreases, and that the unit stress cannot therefore be strictly 
represented by P -^ S, if S is the original cross-section. The 
error is not of practical consequence for this discussion. 

17. Ultimate or Breaking Strength. — If the force ap- 
plied in tension to the bar is increased, a point will next be 
reached where a repeated application of the same foi'ce causes 
a successive increase in the permanent elongation. As this 
phenomenon means a gradual drawing out, final failure by 
pulling asunder is only a matter of a greater or less number 



A PIECE UNDER DIRECT FORCE. II 

of applications of the force. While the bar is apparently 
breaking under this force, the rapid diminution of cross-sec- 
tion near the breaking point actually gives a constantly rising 
unit stress, as will be seen by the dotted curve in the figure. 

If, however, the force is increased without pause from 
the beginning, the breaking force will be higher, as might be 
expected; since much work of deformation is done upon the bar 
before fracture. The bar would have broken under a some- 
what smaller force, applied statically for a considerable time. 

The elongation of the bar was uniform per unit of length 
during the earlier part of the test. There comes a time when 
a portion or section of the bar, from some local cause, begins 
to yield more rapidly thafi the rest. At once the unit stress 
at that section becomes greater than in the rest of the bar, by 
reason of decrease of cross-section, and the drawing out be- 
comes intensified, with the result of a great local elongation 
and necking of the specimen and an assured final fracture at 
that place. If the bar were perfectly homogeneous, and the 
stress uniformly distributed, the bar ought to break at the 
middle of the length, where \}i\Q. flow of the metal is most free. 

It is customary to determine, and to require by specifica- 
tion, in addition to elastic limit and ultimate strength (on one 
continuous application of increasing load), the per cent, of 
elongation after fracture (which is strictly the permanent set) 
in a certain original measured length, usually eight inches, and 
the per cent, of reduction of the original area, after fracture, 
at the point of fracture. As the measured length must in- 
clude the much contracted neck, the avei^age per cent, of 
elongation is given under these conditions. A few inches 
excluding the neck would show less extension and an inch or 
two at the neck would give a far higher per cent, of elonga- 
tion. The area between the axis of X, the extreme ordinate 
and the curve will be the work of fracture, if S is considered 
constant, and will be a measure of the ability of the material 
to resist shocks, blows and vibrations before fracture. It is 
indicative of the toughness or ductility of the material. 

The actual curve described by the autographic attachment 
to a testing machine is represented by the full line; the real 
relation of stress per square inch to the elongation produced, 



12 STRUCTURAL MECHANICS. 

when account is taken of the progressive reduction of sectional 
area, is shown by the dotted Hne. The yield point, or com- 
mon elastic limit, is very marked, there appearing to be a 
decided giving way or rearrangement of the particles at that 
value of stress. The true elastic limit is much below that 
point. The other curves represent the behavior of wrought 
iron as marked. 

i8. Effect of a Varying Cross-Section. — If a test speci- 
men is reduced to a smaller cross-section, by cutting out a 
curved surface, for only a short distance as compared with its 
transverse dimensions, it will show a greater unit breaking 
stress, as the metal does not flow freely, and lateral contrac- 
tion of area is hindered. But, if the portion of reduced 
cross-section joins the rest of the bar by a shoulder, the 
apparent strength is reduced, owing to a concentration of 
stress on the particles at the corner as the unit stress suddenly 
changes from the smaller value on the larger section to the 
greater unit stress on the smaller cross-section. 

19. Compression Curve. — A piece subjected to com- 
pression will shorten, the particles being forced nearer to- 
gether, and the cross-section will increase. It might be 
expected, and is found by experiment to be the case, that, in 
the beginning, the resistance of the particles to approach 
would be like their resistance to separation under tension, so 
that the tension diagram might be prolonged through the 
origin into the third quadrant, reversing the sign of the ordi- 
nate which represents unit stress and of the abscissa which 
shows the corresponding change of length. As this part of 
the diagram is a straight line, it follows that the value of E, 
the elastic modulus for compression, is the same as that for 
tension. After passing the elastic limit the phenomena of 
compression are not so readily determined, as fracture or 
failure by compressive stress is not a simple matter, and the 
increase of sectional area in a short column of ductile material 
will interfere with the experiment. In long columns and with 
materials not ductile, failure takes place in other ways, as will 
be explained later. 

The compression curve is here shown in the same quad- 
rant with the tension curve for convenience and comparison. 



A PIECE UNDER DIRECT FORCE. I3 

It will be seen that, when the tension and compression curves 
are corrected for change of cross-section, they are practically 
the same; and that the resistance per square inch really 
increases up to fracture. 

20. Resilience. — By definition, § lo, if /"is the unit stress 
per square inch and U the stretch of a bar of length /, in 
inches, the modulus of elasticity E = / -^ x, provided f does 
not exceed the elastic limit. Also the work done in stretch- 
ing a bar to that elastic limit by a force P, gradually applied, 
that is, beginning with zero and increasing with the stretch, is 
the product of the mean force, J P, into the stretch, or 

Work done ^ \V)d ^^-^ . fL ^ Jl , S/. 

S/ is the volume of the bar; /^ -^ 2E is called the modu- 
lus of resilience, when /"is the elastic limit, or sometimes the 
maximum safe unit stress. This modulus depends upon the 
quality of the material, and, as it is directly proportional to 
the amount of work that can safely be done upon the bar by 
a load, it is a measure of the capacity of a certain material for 
resisting or absorbing shock and impact without damage. For 
a particular piece, the volume S/ is also a factor as above. 
A light structure will suffer more from sudden or rapid loading 
than will a heavier one of the same material, if proportioned 
for the same unit stress. 

21. Work Done Beyond the Elastic Limit. — The work 
done in stretching a bar to any extent is, in Fig. i, the area 
in the diagram between the curve from the origin up to any 
point, the ordinate to that point and the axis of abscissas, 
provided the ordinate represents P, and the abscissa the total 
stretch. 

Further, it may be seen from the figure that, if a load 
applied to the bar has exceeded the yield point, the bar, in 
afterwards contracting, follows the line D C or F E; and, 
upon a second application of the load, the right triangle of 
which this line is the hypothenuse will be the work done in the 
second application, a smaller quantity than for the first appli- 
cation. But, if the load, in its second and subsequent appli- 
cations, possesses a certain amount of energy, by reason of 



14 STRUCTURAL MECHANICS. 

not being gently or slowly applied, this energy may exceed the 
area of the triangle last referred to, with the result that the 
stress on the particles of the bar may become greater than on 
the first application. Indeed it is conceivable that this load 
may be applied in such a way that the resulting unit stress 
may mount higher and higher with repeated applications of 
load, until the bar is broken with an apparent unit stress 
P -^- S, far less than the ultimate strength, and one which at 
first was not much above the yield point. If the load in its 
first application is above the yield point of the material, and 
it is repeated continuously, rupture will finally occur. 

What is true for tensile stresses is equally true for com- 
pressive stresses, except that the ultimate strength of ductile 
materials under compression is uncertain and rather indefi- 
nite. 

22. Sudden Application of Load. — If a wrought iron or 
soft steel rod, lO feet = 120 inches long, and one square inch 
in section, with E = 28,000,000, is loaded gradually with 
12,000 pounds longitudinal tension, its stretch will be 12,000 
X 120 -h 28,000,000 = 0.05 inches, and the work done in 
stretching the bar will be 6,000 x 0.05 = 300 in. pounds. 

But, if the 12,000 lbs. is suddenly applied, as by the ex- 
tremely rapid loading of a structure of which the rod forms a 
part, or the quick removal of a support which held this weight 
at the lower end of the rod, the load will cause a greater stretch 
at first, after which the rod will contract and then undergo a 
series of longitudinal vibrations of decreasing amplitudes, fin- 
ally settling down to a stretch of 0.05 inches when the extra 
work of acceleration has been absorbed. 

To ascertain what suddenly applied force will produce, at 
the most, a stretch of 0.05 in., and hence the same unit stress 
as a quiescent load (for stretch and stress are directly related), 
observe that the work done by the application of a gradually 
applied load is represented by the area of a triangle of 300 
in. pounds, the force increasing from zero to P, with a mean 
value of J P. A constant force during the stretch gives a rec- 
tangle for the area representing work done, and can be only 
300 -4-0.05 = 6,000 lbs., or half as much as before. The 
work of acceleration on the mass of the bar is neglected. 



A PIECE UNDER DIRECT FORCE. 



15 



Stresses produced by moving loads on a structure are in- 
termediate in effect between these two extremes, depending 
upon rapidity or suddenness of loading. Hence it is seen why 
the practice has arisen of limiting stresses due to moving loads 
apparently to only one-half of the values permitted for those 
caused b}^ static loading. 

For resilience or work done in deflection of beams, see 

§ 114- 

23. Granular Substances Under Compression. — Fail- 
ure by Shearing on Oblique Planes. Blocks of material, such 
as cast-iron, sandstone or concrete, when subjected to com- 
pression, frequently give way by fracturing on one or more 
oblique planes which cut the block into two wedges, or into 
pyramids and wedges. The pyra- 
mids may overlap, and their bases 
are in the upper and lower faces of 
the block. This mode of fracture, 
peculiar to granular substances, of 
comparatively low shearing resist- 
ance, can be discussed as follows: 

If a short column, Fig. 2, of cross- 
section S is loaded centrally with P, 
the unit compression on the right 
section will be /i = P ^ S, and, if 
the short column gives way under this load, this value of p^ is 
commonly considered the ■ crushing strength of the material. 
While it doubtless is the available crushing strength of this 
specimen, it may by no means represent the maximum resist- 
ance to crushing under other conditions. 

If /i = P ^ S is the unit thrust on the right section, it 
is seen, from § 180, that, on a plane making an angle d with 
the right section, the normal unit stress, /„ = p^ cos^ Q, and 
the tangential unit stress q = p^ sin 6 cos d. If m = co- 
efficient of frictional resistance of the material to sliding, the 
resistance per square inch to sliding along this oblique plane 
will be mp,, = mpi cos^ ^, and the portion of the unit shear- 
ing stress tending to produce fracture along this plane will be 
q — mpn = pi (sin d cos — m cos^ d). 

Fracture by shearing, if it occurs, will take place along 




Fi^.2. 



1 6 STRUCTURAL MECHANICS. 

that plane for which the above expression is a maximum, or 
d{q — i)ip») -^ d = o. Differentiating relatively to ^, 



/j (cos^ — sin'-^ -\- 2 7n sin cos <9) = o. 

Siir — 2 m sin cos -{- vv- cos^ = cos''^ Q -}- trC" cos'^ Q; 

sin ^ — ;;^ cos = cos 6^ -|/(i -j" ^^^0- 

sin <9 , /^ o 

~ = tan 6* = ;;z + -i/(i -I- m^). 

cos ^ I I V I y 

If m were zero, ^ max. would be 45°. Therefore the plane 
of fracture always makes an angle greater than 45° with the 
right section. As 6 may be negative as well as positive, frac- 
ture tends to form pyramids or cones. 

Example. — A rectangular prism of cast-iron, 2 in. high and 
square section = 1.05 sq. in., sheared off under a load of 97,000 
lbs., or 92,380 lbs. compression per sq. in. of cross-section, at an 
angle whose tangent was 1.5, or 56° 19'. 

1.5 = m -j- 1/(1 + m^')) 2.25 — y?i -j- 111^ =: i -)- m^-^ m = 0.42. 

Sin 6 — 0.8921, cos 6 = 0.5546, sin d cos 6 = 0.495, cos^ d = 0.308. 

92,380(0.495 — 0.42 X 0.308) = 33.800 lbs. 

The coefficient of friction is 0.42, and the shear 33,800 lbs. per sq. 
in. The crushing strength of a short block would have exceeded 
considerably the above 92,400 lbs. 

Since this deviation of the plane of fracture from 45° is 
due to a resistance analogous to friction, it follows that, 
when a column of granular material, and of moderate length, 
gives way by shearing, the value p^ will be only that com- 
pressive stress which is compatible with the unit shearing 
strength, while its real compressive strength in large blocks 
will be much higher. 

The same phenomenon is exhibited by blocks of sand- 
stone and of concrete. Tests of cubes and flat pieces yield 
higher results than • do those of prisms of the same cross- 
section and having a considerably greater height. See plate 
I. for views of failure by shearing of cast-iron and sandstone. 
24. Ductile Substances Under Compression. — Wrought 
iron, and soft and medium steel, as well as other ductile sub- 
stances, tested in short blocks in compression, bulge or swell 
in transverse dimensions, and do not fracture. Hence the 
ultimate compressive strength is indefinite. 



A PIECE UNDER DIRECT FORCE. I 7 

25. Fibrous Substances Under Compression. — Wood 
and fibrous substances which have but small lateral cohesion 
of the fibres, when compressed in short pieces in the direction 
of the same, separate into component fibres at some irregular 
section, and the several fibres fail laterally and crush. See 
Plate I. for crushing of wood. 

26. Vitreous Substances Under Compression. — Vitre- 
ous substances, like glass and vitrified bricks, tend to split in 
the direction of the applied force. 

27. Resistance of Large Blocks. — The resistance per 
square inch of a cube to compression will depend upon the size of 
the cube. As the unit stress and the resulting deformation are asso- 
ciated, as noted in § 4, it follows that the unit compressive stress 
will be greatest at the centre of the compressed surface and least 
at the free edges where lateral movement of the particles is 
less restrained. Hence, the larger the cube, the greater the mean 
or apparent strength per square inch. Large blocks of stone, 
therefore, have a greater average sustaining power per square inch 
than is indicated by small test specimens, other things being equal. 

The same inference can be drawn as to resistance of short 
pieces to tension as compared with longer pieces of the same 
cross-section. 

A uniform compression over any cross-section of a large post 
or masonry pier, when the load is centrally applied to but a small 
portion of the top can be realized only approximately; the same thing 
is probably true of the foundation below the pier. The resisting 
capacity of the material, if earth, is thereby enhanced; for the 
tendency to escape laterally at the edges of the foundation is not 
so great as would be the case if the load were equally severe over 
the whole base. 

Beveling the edges of the compressed face of a block will 
increase the apparent resistance of the material by taking the load 
from the part least able to stand the pressure. The unloaded 
perimeter may then act like a hoop to the remainder. 

Examples. — i. A round bar, 1 in. in diameter and 10 ft. long 
stretches 0.06 in., under a pull of 10,000 lbs. What is the value 
of E? What is the work done? 25,464,733; 300 in. pounds. 

2. If the elastic limit of the bar is reached by a tension of 
30,000 lbs. per sq. in., what is the work done or the resilience of 
the bar? 1,666 in. pounds. 

3. An iron rod, E = 29,000,000, hangs in a shaft 1,500 ft. 
deep. What will be the stretch? 1.55 in. 

4. A certain rod, 22 ft. long, and having E = 28,000,000, is 
to be adjusted by a nut of 8 threads to the inch to an initial ten- 
sion of 10,000 lbs. per sq. in. If the connections were rigid, how 



l8 STRUCTURAL MECHANICS. 

much of a revolution ought to be given to the nut after it fairly 
bears? o-75- 

5. Can a weight of 20,000 lbs. be lifted by cooling a steel 
bar, I in. sq. from 212'^ to 62° F.? Co-efficient of expansion 
= .0012 for t8o°; E = 29,000,000. 

6. A steel eyebar, 80 in. long and 2 in. sq., fits on a pin at 
each end with 1-50 in. play. What will be the tension in the bar, 
if the temperature falls 75° F. and the pins do not yield? 

7,250 lbs. per sq. in. 

7. A cross grained stick of pine, one sq. in. in section 
sheared off at an angle of about 66° with the right section under a 
compressive load of 3,200 lbs. If the coefficient of friction is 
0.5, what is the unit shearing stress of the section, the actual 
irregular area being 2.9 sq. in. ? 800 lbs. 



CHAPTER II. 

V MATERIALS . 

28. Growth of Trees. — Trees from which himber is cut 
grow by the formation of woody fibre between the trunk and 
the bark, and each annual addition is more or less distinctly 
visible as a ring. The sap circulates through the newer wood, 
and in most trees the heart wood, as it is called, can be easily 
distinguished from the sap wood. The former is considered 
more strong and durable, unless the tree has passed its prime. 
The heart then deteriorates. Sap wood, in timber exposed 
to the weather, is the first to decay. 

Branches increase in size by the addition of rings, as does 
the trunk; hence a knot is formed at the junction of the branch 
with the trunk. The knot begins where the original bud 
started, and increases in diameter towards the exterior of the 
trunk, as the branch grows. The grain of the annual growth, 
formed around the junction of the branch with the trunk, is 
much distorted. Hence timber that contains large knots is 
very much weaker than straight grained timber. Even small 
knots determine the point of fracture when timber is experi- 
mentally tested for strength. When a branch happens to die, 
but the stub remains, and annual rings are added to the trunk, 
a dead or loose knot occurs in the sawed timber; such a knot 
is considered a defect, as likely to let in moisture and start 
decay. 

As forest trees grow close together, the branches die suc- 
cessively from below from lack of sunlight; such trees develop 
straight trunks of but little taper, free from any knots, except 
insignificant ones immediately around the centre, and yield 
straight grained, clear lumber. A few trees, like hemlock, 
sometimes have their fibres running in a spiral, and hence 
yield cross-grained timber. Trees that grow in open spaces 
have large side limbs, and the lumber cut from them 
has large knots. 



20 STRUCTURAL MECHANICS. 

29. Shrinkage of Timber. — If a log is stripped of its 
bark and allowed to dry or season, it will be found that the 
contraction or shrinkage in the direction of the radius is prac- 
tically nothing. There are numerous bundles or ribbons of 
hard tissue running radially through the annual rings which 
appear to prevent such shrinkage. Radial cracks, running in 
to a greater or less distance, indicate that the several rings 
have yielded to the tension set up by the tendency to shrink 
circumferentially. Sawed timber of any size is likely to ex- 
hibit these season cracks. Such cracks are blemishes and may 
weaken the timber when used for columns or beams. By 
slow drying, and by boring a hole through the axis to promote 
drying withm, the tendency to form season cracks may be 
diminished. 

A board sawed radially from a log will not shrink in width, 
and will resist wear in a floor. Such lumber is known as 
quaj'ter sawed. A board taken off near the slab will shrink 
much and will tend to warp or become concave on that side 
which faced the exterior of the log. For that reason, and 
because the annual rings have less adhesion than the individ- 
ual fibres have, all boards exposed to wear, as in floors, should 
be laid heart-side down. 

30. Decay of Wood. — Timber exposed to the weather 
should be so framed together, if possible, that water will not 
collect in joints and mortises, and that air may have ready 
access to all parts, to promote rapid drying after rain. The 
end of the grain should not be exposed to the direct entrance 
of water, but should be covered, or so sloped that water can 
run off, and the ends should be stopped with paint. It is well 
to paint joints before they are put together. 

The decay of timber is due to the presence and action of 
vegetable growths or fungi, the spores of which find lodgment 
in the pores of the wood, but require air and moisture, with a 
suitable temperature, for their germination and spread. 
Hence if timber is kept perfectly dry it will last indefinitely. 
If it is entirely immersed in water, it will also endure, as air 
is excluded. Moisture may be excluded from an exposed sur- 
face by the use of paint. Unseasoned timber placed where 
there is no circulation of air will dry-rot rapidly in the interior 



MATERIALS. 2 1 

of the stick; but the exterior shell will be preserved, since it 
dries out or seasons to a little depth very soon. 

The worst location for timber is at or near the ground 
surface; it is then continually damp and rot spreads fast. 

31. Preservation of Wood. — The artificial treatment of 
timber to guard against decay may be briefly described as the 
introduction into the pores of some poison or antiseptic to 
prevent the germination of the spores; such treatment is effi- 
cacious as long as the substance introduced remains in the 
wood. 

The most common fluids forced by pressure into the 
pores of the wood are cresote and zinc chloride; although cor- 
rosive sublimate, copper sulphate and other chemicals have 
been employed. As much as from 12 to 20 pounds of creo- 
sote or dead oil of tar have been forced into a cubic foot of 
timber, adding materially to its weight. From timber partly 
immersed in water the creosote may wash out to some 
extent. 

Burnettizing is the name given to treatment with zinc 
chloride, a comparatively cheap process, applied to railway 
ties, paving blocks and bridge timbers. To prevent the zinc 
chloride from dissolving out in wet situations, tannin has been 
added after the zinc, to form with the vegetable albumen a 
sort of artificial leather, plugging up the pores. Hence the 
name, zinc-tannin process. 

32. Strength of Timber. — The properties and strength 
of different pieces of timber, classed as belonging to the same 
species, are very variable. Names differ in different parts of 
the country. The denser and heavier timber is generally the 
stronger, and seasoned is stronger than green lumber. Pru- 
dence would dictate that structures should be designed for the 
strength of green or moderately seasoned timber of average 
quality. As the common woods have a comparatively low 
resistance to compression across the grain, particular attention 
should be paid to providing sufficient bearing area where a 
strut or post abuts on the side of another timber. An inden- 
tation of the wood destroys the fibre and increases the liability 
to decay, if the timber is exposed to the weather, especially 
under the continued working produced by moving loads. 



22 



STRUCTURAL MECHANICS. 



The average breaking stresses of the common woods may 
be stated as follows, in thousands of pounds per square inch. 



White Oak 

Southern Long Leaf or Georgia Pine. 

Douglas or Oregon Fir or Pine 

Northern or Short Leaf Yellow Pine. 

Norway Pine 

Spruce 

White Pine 

California Red wood '. 



TENSION 


COMPRESSION 


BENDING OR 
TRANSVERSE 


z 

< 

Bi 

X 
H 

? 

< 

U 
S 


< 

o 

a: 

H 

10 


6 

Q 

M 

o 

a: 
o 
< 


Z 

< 
ft! 



I 


d 

Q 
in 

CO 

O 

o 
< 


M (^ td 

W M K 

Id 


D c/i 

o ►J 


2 


7 


2 


6 


1, 100,000 


.8 


12 


.6 


8 


1.4 


7 


1,600,000 


.6 


12 




8 


1.2 


6-5 


1,400,000 


.6 


9 


•5 


6 


I 


6 


T, 200,000 


•4 


8 




6 


.8 


4 


1,200,000 




8 


•5 


6 


• 7 


4 


1,200,000 


•4 


7 


•5 


5-5 


.8 


4 


1,000,000 


•4 


7 






.8 


4-5 


700,000 


•4 



Timber has no well-defined yield point. 

33. Timber Specifications. — The following is a specifi- 
cation for bridge or trestle timber: — All timber shall be of the 
best quality of white pine, long leaf yellow pine, or white oak, 
sawed true to size and out of wind (z. e., without twist). It 
shall be free from sap wood, except in sticks having a depth 
of 16 inches or upward, when one inch of sap will be allowed 
on two corners. It shall be free from wind shakes, large 
knots, loose or rotten knots, wormholes, decayed wood, or 
other defects that will impair its strength or durability. 

Sometimes sap wood is not proscribed, and, in that case, 
in order that bark may not show on any corner, the require- 
ment is introduced, no wane. 

Timber sawed from dead trees is very deficient in strength. 

34. Iron and Carbon. — From the standpoint of the per- 
son who uses cast-iron, wrought iron and steel, these materials 
differ from one another in physical qualities on account of the 
different percentages of carbon in combination with the iron. 
The proportion of this ingredient may range from perhaps 
five per cent, to zero, although a portion of the carbon may 
be replaced by some other element. 

35. Cast-iron. — Cast-iron contains the largest percent- 
age of carbon, from two to five per cent., which it gets from 
the fuel. The ore, fuel, and limestone, the latter added as a 
flux for the earthy ingredients, are introduced together at the 
top of a blast furnace, and the molten iron is drawn off at 



MATERIALS. 23 

the bottom, and run into grooves in the sand, the process 
being continuous. This product is known as pig or cast iron. 
Although the slag floats on the melted iron, the cast pig, 
which has taken up from the burning fuel all the carbon it has 
an affinity for, contains some slag and other impurities. 
When broken, it is seen to be crystalline in appearance, and 
it differs in grade from white to gray cast-iron, according to 
the temperature of the furnace and abundance of fuel. Gray 
cast-iron is more fluid when melted than white iron, but it 
requires a higher temperature for its fusion. Gray cast-iron 
contains one per cent., and sometimes less, of carbon in 
chemical combination with the iron, and from one to tJiree or 
four per cent, of carbon, in the state of graphite, in mechan- 
ical mixture; while white cast-iron is a chemical compound of 
iron with from two to four per cent, of carbon. The graphite 
can be brushed from a freshly broken surface of gray iron, 
and can have no effect on the metal, except to diminish its 
strength by preventing a complete union of particles. 

The melting point falls, and the hardness and brittleness 
increase, with the increase of carbon in chemical combina- 
tion. White cast-iron is of no practical use in itself, but is 
used in making wrought iron and steel. Gray iron is used in 
the arts, although its brittleness often renders it objectionable. 
The softest grade is probably the most fluid, and, since it 
flows well into the mold, is used for making thin castings, as 
in ornamental cast-iron work, where strength is not required. 

An intermediate grade can be converted from white cast- 
iron to gray iron by fusion and slow cooling, the carbon hav- 
ing time to separate, and from gray to white by fusion and 
sudden cooling. When such melted metal is run into a mold 
lined with iron, it is chilled, from the surface for a depth of 
from one-half to three-fourths of an inch, and made intensely 
hard, as in the treads of car-wheels. Note the parallelism to 
the hardening of steel. The tenacity of cast-iron may range 
from 15,000 lbs. to 30,000 lbs. per square inch of section, and 
its compressive strength from 60,000 lbs. to 100,000 lbs. Its 
modulus of elasticity is from 17 to 20 millions. So crude a 
product is improved by successive remeltings; hence old cast- 
ings can be melted over with advantage. 



24 STRUCTURAL MECHANICS. 

36. Wrought Iron. — Direct processes for making 
wrought iron or steel from the ore may be employed, but they 
are wasteful of iron. Commonly, white pig iron is melted in 
a cupola furnace and then run into a Bessemer converter or a 
reverberatory furnace for treatment, which consists principally 
in the exposure to currents of air to burn out the excess of 
carbon. Other impurities such as silicon, sulphur and phos- 
phorus may at the same time and by the same means be 
reduced in amount or burned out. 

To explain why wrought iron bars are fibrous: — In the 
puddling furnace the surface of the melted pig-iron is exposed 
to a blast rich in oxygen, and the puddler stirs the mass to 
hasten the burning out of the carbon. As the carbon is 
removed, the melting point rises, and the iron becomes thick 
or pasty. Cast-iron does not take on this intermediate state 
between fluid and solid, but wrought iron does, and hence can 
be welded. The semi-fluid iron is collected into a lump by 
the puddler and withdrawn from the furnace. It is then much 
like a sponge; the particles of wrought iron have adhered to 
one another, but each particle of iron is more or less coated 
with a thin film of slag and oxide, as water is spread through 
the pores of a partly dry sponge. 

The lump of iron is put into a squeezer, and the fluid 
slag and oxide drip out as water does from a squeezed sponge. 
But, as it is impracticable to squeeze a sponge perfectly dry, 
so it is impracticable to squeeze all the impurities out from 
among the particles of metallic iron. In the subsequent pro- 
cesses of rolling and re-rolling, each globule of iron is elong- 
ated, but the slag and oxide are still there; so that the rolled 
bar consists of a collection of threads of iron, the adhesion 
of which to each other is not so great as the strength of the 
threads. 

If the surface of an iron bar is planed smooth and then 
etched with acid, the metal is dissolved from the surface and 
the black lines of impurities are left distinctly visible. 

That wrought iron is fibrous is then an accident of the 
process of manufacture, and does not add to its strength. If 
these impurities had not been in the iron when it was rolled 
out, it would have been more homogeneous and stronger. 



MATERIALS. 25 

The fibrous fracture of a bar which is nicked on one side and 
broken by bending is not especially indicative of toughness; 
for soft steel is tough and ductile without being fibrous. 

If the iron first rolled is twice cut up into pieces, piled, 
reheated and rolled, it makes double refined iron, the grade 
used for bridges, and superior to single refined iron, or mer- 
chant bars. The tensile strength of the former is about 
50,000 lbs., varying with the size of the bars; that is, the 
more work done in rolling, the stronger the iron. The com- 
pressive strength is rather low, owing to the ductility of the 
metal, a pressure of from 36,000 lbs. to 40,000 lbs. per 
square inch causing decided lateral flow or bending. E = 28 
or 29,000,000. 

Latterly, the manufacture of what is known as soft steel, 
or homogeneous metal, has been brought to such perfection 
that steel competes in price with wrought iron and has largely 
driven it out of the market. 

37. Steel. — Steel is made from pig-iron by the Bessemer, 
open-hearth, and other processes, all of which have for their 
main object the burning out of the carbon, either completely 
or very nearly. The process is a comparatively rapid one, 
and several tons are treated at once. The heat generated by 
the union of the carbon with the oxygen of the air is sufficient 
to keep the mass fluid, although the melting point rises. 

If the product, when practically freed from carbon, is 
run into molds, and the resulting ingots are rolled out into 
bars and plates, the material is known as soft or mild steel, 
ingot iron or homogeneous metal, but is only iron freed from 
•carbon; it is fine grained, tough and ductile, purer and hence 
stronger than wrought iron. Such a rolled product is not 
fibrous, like puddled iron, but it is practically the same mate- 
rial, of a better quality. It will weld, but will not harden and 
temper. 

While the reduction of the amount of carbon in the metal 
under treatment in the furnace or converter may be stopped 
at a certain desired point, the carbon may be all removed, and 
then the proper amount may be added to the charge to pro- 
duce steel with a certain percentage of carbon. By this means 
the character of the product is better assured. The carbon is 



26 STRUCTURAL MECHANICS. 

often added in the form of a certain iron, previously melted, 
of definite composition, known as spiegel-eisen. It also con- 
tains manganese, which is beneficial in helping to eliminate 
sulphur. The product of the Bessemer process, very exten- 
sively used for steel rails, is not usually so uniform in charac- 
ter as steel made by the open hearth process, where time is- 
afforded for testing the metal and then bringing it to the de- 
sired grade. 

A very small amount of sulphur in steel or iron will ren- 
der it red-short, that is brittle and liable to check and crack 
when hot enough to roll and forge. Phosphorus renders the 
metal cold-short, or brittle at ordinary temperatures. Cur- 
rent specifications will show how very little of either is toler- 
ated in the best steel. 

Since steel is iron with a very small percentage of carbon, 
it may be made by melting wrought-iron with carbon in a cru- 
cible; when the iron takes up carbon, and the product is 
known as crucible steel. This steel, when rolled, sheared up 
and rolled again, is shear steel, from which cutting tools are 
made. Such steel contains from o. 5 to i . 5 per cent, of carbon. 

The common soft steel which is used for tension mem- 
bers of bridges, and for pieces exposed to violent use, shocks 
and vibrations, does not probably contain more than o. 10 to 
o. 12 of one per cent, of carbon. Steel for compression mem- 
bers, known as medium steel, will contain from 0.12 to 0.24 1 
or o. 36 of one per cent. Thus the carbon in structural steel 
will range from zero to one-third of one per cent. , a range be- 
low the o. 5 of one per cent., for steel which hardens and tem- 
pers. Practically the steel used for structures is nothing more 
than a better grade of iron under a different name. 

Steel, properly so-called, will harden and temper, and 
will not weld; but the use of the electric arc permits the ends 
of two pieces to be melted together. As the percentage of 
carbon falls off. steel loses the property of hardening and 
tempering, and takes on the property of welding. The chill- 
ing of a certain grade of cast-iron is analogous to the harden- 
ing ot steel. The tensile strength of the softest steel runs 
from 50,000 lbs. to 60,000 lbs. per sq. inch. This steel is 
also used for boilers. The next or medium grade of steel will 



i 



MATERIALS. 2 J 

run from 60,000 lbs. to 70,000 lbs., and the steel which is 
used for columns may resist at failure from 70,000 lbs. to 
80,000 lbs. per sq. inch. Tool steel will go higher. E = 28 
to 30,000,000. 

The tensile strength of cast-iron, wrought iron and 
medium steel may be roughly represanted by the series i, 2 
and 3, corresponding to 25,000, 50,000 and 75,000 pounds 
per square inch. 

Article's cast from steel are taking the place of iron 
castings where great strength and toughness are required. 
They are superior to and cheaper than common forged 
work. § 47. 

To recapitulate: — Wrought or puddled iron is fibrous by 
reason of the way in which it is made; soft steel is fine grained; 
the hardest steel is crystalline; cast-iron is coarsely granular. 

Cast-iron contains from two to five per cent, of carbon, 
a part in chemical combination, the rest in mechanical mix- 
ture as graphite. Cast-iron contains the most carbon; steel, 
such as is used for machine and other tools, has a medium 
anount of carbon; the soft, structural steel has very little, and 
some of it, practically, none at all; and wrought iron has no 
carbon. 

38. Malleable Iron: Case-hardened Iron. — There are 
two other products which may well be mentioned, and which 
will be seen to unite or fit in between the three already 
described. The first is what is known as "malleable cast- 
iron" or malleable iron. 

Small articles, thin and of irregular shapes, which may 
be more readily cast than forg"ed or fashioned by a machine, 
and which need not be very strong, are made of cast-iron, 
and then imbedded in a substance rich in oxygen, as, for 
instance, powdered red hematite iron ore, sealed up in an iron 
box, and heated to a high temperature for some time. The 
oxygen abstracts the carbon from the metal to a shght depth, 
converting the exterior into soft iron, malleable iron, with an 
increase of strength and diminution of brittleness. 

The second product is case-hardened iron. An article 
fashioned of wrought iron or soft steel is buried in powdered 
charcoal and heated. The exterior absorbs carbon and is 



28 



STRUCTURAL MECHANICS. 



converted into high steel, which will better resist wear and 
violence than will soft iron. The Harvey process for harden- 
ing the exterior of steel armor plates is of a similar nature. 

39. Work of Elongation. — It is seen from the diagram, 
Fig. I, that the resistance of the metal per square inch increases 
as the bar draws out and diminishes in section under tension, as 
shown by the dotted curve, although the total resistance grows 
less near the close of the test, as shown by the full line. As a 
small increase in the amount of carbon diminishes the elongation 
and reduction of area, it is possible that the carbon affects the 
apparent ultimate strength in this manner (since such strength 
is computed on the square inch of original section), and not by 
actually raising the resisting power of the metal. 

Since the measure of the work done in stretching a bar is the 
product of one-half the force by the stretch, if the yield point has 
not been passed, and, for values beyond that point, is the area 
below the curve in the diagram, limited by the ordinate represent- 
ing the maximum force, — the comparative ability of a material to 
resist live load, shock and vibration is indicated by this area. A 
mild steel of moderate strength may thus have greater value than 
a higher carbon steel of much greater tensile strength. 

Such a measure of work done to produce fracture is aimed 
at in the following specification: 

In tensile tests of steel, round sample bars, not less than |- 
in. in diameter, and not less than 12 diameters long between the 
jaws of the testing machine, shall show a percentage of elongation 
not less than 1,200,000 -^ the tensile strength per square inch, and 
a percentage of reduction of area not less than double the same 
ratio. The elongation shall be measured after breaking, on an 
original length of ten times the diameter of the piece, in which 
length must occur the curye of reduction from stretch on both 
sides of the point of the fracture. 

40. Classification. — La Societe Cockerill has proposed the 
following classification for steel: — 

Elonga- 
Breaking Strength, tion, ^ 



Steel. 
Extra Soft 



Carbon, %. 
0.05 to 0.20 

Soft 0.20 to 0.35 

Hard 0.35100.50 

Extra Hard 0.50 to 0.65 



57,000 to 70,000 271020 

70,000 to 85,000 20 to 15 

85,000 to 100,000 15 to 10 

100,000 to 115,000 10 to 5 



\ Welds but does not 
( temper. 

( Welds with difficulty; 
/ does not temper, 
i Does not weld. Tem- 
} pers. 

Does not weld. Tem- 
pers hard. 



The first and second grades would more commonly be called 
so/^ and mediu77i. 

From recent experiments it appears that wrought iron, or 
steel almost free from carbon, can be hardened, if heated hot 



MATERIALS. 29 

enough before sudden cooling, and that a great increase in strength 
is thus developed. This fact tends to confirm the view that 
increased strength is not due to carbon percentage but to dimin- 
ished lateral contraction under longitudinal tension. See § 200. 

Iron may also be increased in softness by very carefully 
annealing. 

41. Specifications for Cast-iron. — All castings must be 
of tough, gray iron, free from cold shuts or injurious blow 
holes, true to form and thickness, and of a workmanlike fin- 
ish. Sample pieces, one inch square, cast from the same heat 
of metal in sand molds, shall be capable of sustaining, on a 
clear span of 4 ft. 6 in., a central load of 500 lbs. , when tested 
in the rough bar. A blow from a hammer shall produce an 
indentation on a rectangular edge of the casting without flak- 
ing the metal. 

Cast-iron for water pipes is usually required to have from 
16,000 to 18,000 lbs. per sq. in. tensile strength, and to be 
soft enough to admit readily of tapping and cutting a thread. 

42. Specifications for Wrought Iron. — Wrought iron 
has been to a great degree displaced by mild or soft steel. 
The following specifications for wrought iron agree very well 
with recent practice: — 

All wrought iron must be tough, fibrous and uniform in 
character. It shall have a yield point of not less than 26,000 
lbs. per sq. in. Finished bars must be thoroughly welded dur- 
ing rolling, and be free from injurious seams, blisters, buck- 
les, cinder-spots, or imperfect edges. 

For all tension members, the muck bars shall be rolled 
into fiats, and again cut, piled and rolled into finished sizes. 
They shall stand the following tests: — Full-sized pieces of fiat, 
round or square iron, not over 4J square inches in sectional 
area, shall have an ultimate strength of 50,000 lbs. per sq. 
in., and stretch I2i per cent, in their whole length. 

Bars of larger section than 4J square inches, when tested 
in the usual way will be allowed a reduction of 1,000 lbs. for 
each additional square inch, down to a minimum of 46,000 
lbs. per square inch. 

When tested in specimens, of uniform section of at least 
one-half a square inch in a length of ten inches, taken from ten- 
sion members rolled to a section not more than 4I square 



30 STRUCTURAL MECHANICS. 

inches, thfe iron shall show an ultimate strength of 52,000 lbs., 
and stretch 18 per cent, in a measured distance of eight 
inches. Specimens from bars larger than 4^ square inches 
will be allowed a reduction of 500 lbs. for each additional 
square inch of section, down to minimum of 50,000 lbs. 

The same sized specimen taken from angle and other 
shaped iron, or from plate iron, shall have an ultimate strength 
of 50,000 lbs., and elongate 15 per cent, in eight inches. 

All iron for tension members must bend cold for about 
90°, to a curve of diameter not over twice the thickness of the 
piece, without cracking; and at least one sample in three must 
bend 180° to this curve without cracking. When nicked on 
one side and bent from a blow with a sledge, the fracture 
must be nearly all fibrous, showing but few crystalline specks. 
Specimens from angle, plate and shaped iron must stand bend- 
ing cold through 90°, to a curve of diameter not over three 
times the thickness of the piece, without cracking. When 
nicked and bent, the fracture of the specimen must be mostly 
fibrous. 

43. Specifications for Steel. — The following specifica- 
tions have been used for structural steel: — 

Steel for rivets and eyebars shall contain not more than 
one quarter of one per cent, of carbon, and less than one tenth 
of one per cent, of phosphorus. A sample bar | in. in dia- 
meter, when tested in a lever machine, shall have a yield 
point of not less than 40,000 lbs. per square inch, and an ul- 
timate strength of not less than 70,000 lbs. per square inch; 
it shall elongate at least 18 per cent, in a length of eight 
inches, and shall show a reduction of area of at least 45 per 
cent, at the point of fracture. In full-sized bars this steel 
shall have a yield point of at least 35,000 lbs., and an ultimate 
strength of at least 65,000 lbs. per square inch; it shall elon- 
gate 10 per cent, before breaking, and for stresses less than 
30.000 lbs. per square inch, it shall show a modulus of elas- 
ticity between 28,000,000 and 30,000,000. 

A sample bar | inch in diameter shall bend 180° cold, and 
be set back upon itself without showing crack or flaw. 

Steel used in compression members shall not contain 
more than one-tenth of one per cent, of phosphorus. When 



MATERIALS. 3 1 

tested in tension, a sample bar | in. in diameter shall have an 
ultimate strength of not less than 80,000 lbs. per square inch, 
and'' a yield point of not less than 50,000 lbs; it shall elongate 
at least i 5 per cent, in eight inches, and show a reduction of 
area of at least 30 per cent, at the point of fracture. It shall 
also bend 180° cold around its own diameter without showing 
crack or flaw. It shall be incapable of tempering. The fol- 
lowing are examples of steel, not annealed: — 



Yield Ultimate Elongation Reduced ° tic ^ 

Point. Strength. in 8 in. Area. *- S 2 

u S CL, 

53,i4olbs. 83,6Solbs. 20.75% 37-78% 0.27 0.83 .067 ) of one 

5i,i9olbs. 84,44olbs. 18.75% 33-23% 0.26 0.79 .067 j" per cent 

For structural and architectural work the following re- 
quirements are reasonable: 

Rolled steel shall not contain more than 0.04 per cent, 
of phosphorus. It shall be finished straight and smooth. 
It shall show by the standard test piece an ultimate tensile 
strength of 56,000 ± 4,000 lbs. per sq. inch, with an elonga- 
tion of 25 per cent, in eight inches. The yield point shall in 
no case be less than 55 per cent, of the maximum tension 
sustained by the test piece (.55 . 56,000 = 30,800), and it 
may be indicated by the "drop of beam" of the testing 
machine. Each melt of finished material shall receive three 
tests: two tension (one cut from each extreme variation in 
thickness of metal), and one hot bending. When the chemi- 
cal analysis shows less than 0,05 per cent, of sulphur, the hot 
bending test will not be required. 

44. Machinery Steel. — When much machine work is 
to be done on a piece, use for forgings mild steel of from 
0.20 to 0.25 of one per cent, of carbon, with a yield point of 
27,000 lbs. per sq. inch, an ultimate strength of 57,000 lbs. 
per sq. inch, and an average elongation of 28 per cent, in a 
length of four diameters. 

For the general run of engine forgings use a harder steel 
of 35,000 lbs. per sq. inch yield point, an ultimate strength 
of 75,000 lbs., and having 25 per cent, elongation in four 
diameters. 



32 STRUCTURAL MECHANICS. 

With precautions, forgings may be made from steel of a 
still higher grade, for crank and crosshead pins, and parts 
subjected to severe alternating stresses and wearing action, 
such steel to show a yield point of 40,000 lbs. per sq. inch, 
an ultimate strength of 85,000 lbs., and 20 per cent, elonga- 
tion in four diameters. 

If the shape of the piece will allow tempering, the above 
values may be raised to a yield point of 45,000 lbs., an ulti- 
mate strength of from 85,000 to 90,000 lbs., and 23 percent, 
elongation. 

A small percentage of nickel added to mild steel increases 
the strength greatly without causing loss of ductility. The 
hollow shafts for the U. S. S. Brooklyn, 17 inches external 
diameter, 1 1 inches internal diameter, 39 feet long, showed 
60,775 lt)S. yield point, 94,245 lbs. ultimate strength, and 
60. 5 per cent, reduction of area. 

Fluid compressed, oil-tempered steel, containing from 0.4 
to 0.45 of one per cent, of carbon, and especially nickel-steel, 
is suitable for piston-rods of rock-drills, hammer rods, stamp 
stems, cam shafts, crank and crosshead pins, and pieces sub- 
jected to alternating stresses of tension and compression, or 
to either kind frequently repeated. The yield point will be 
from 50,000 to 60,000 lbs. 

Steel castings are now successfully made, although form- 
erly much difficulty was experienced in securing soundness. 
Some have shown, before annealing, a tensile strength of 
90,000 lbs. per square inch, an elongation of 22%, and a 
reduction of area of from 30 to 40%. 

45. Chemical Specifications and Manipulation. — It is 
the opinion of some engineers that neither the chemical con- 
stitution nor the mechanical processes of manufacture should 
be specified, in calling for a certain grade of steel, but only 
breaking strength, yield point, elongation and reduction of 
area. The following, however, are examples of requirements 
in some specifications: — 

Acid open-hearth and Bessemer steel limits for phosphorus 
may range from 0.06 to o.i of one per cent., and for sulphur one- 
half of these percentages. Basic open-hearth steel may have from 
0.02 to 0.06 of one per cent, of phosphorus, and from 0.02 to 0.04 
of one per cent, of sulphur. 



MATERIALS. 33 

Structural Steel: — To be acid or basic open-hearth. Acid 
steel to have phosphorus below 0.06%; basic, below 0.04%. Sul- 
phur to be below o.io; silicon, below o.io; manganese, below 
0.65. To show 56,000 to 64,000 lbs. ultimate strength; ratio of 
yield point to ultimate strength, 50 to 63%; elongation, 2 7 to 
22%; and reduction of area 50 to 40%. 

Another specification for structural steel for railroad 
bridges says: — 

All raw material used in the manufacture of steel ingots shall 
be chemically within the Bessemer limit of phosphorus o.io of one 
per cent., sulphur 0.05, copper 0.40. 

All ingots shall be cast from steel melted in an acid-lined 
open-hearth furnace. No single ingot or casting shall exceed 
15,000 lbs. in weight, in order to avoid extreme segregation. All 
ingots must be bottom cast, and no ingot shall be disturbed or 
removed from the position in which it is cast until it is sufficiently 
solidified to obviate ''bleeding." 

Finished rolled steel shall show under analysis not more than 
0.08% phosphorus, 0.04 sulphur, 0.45 manganese, 0.20 copper. 

It shall be straight, well finished in the rolling, full to 
dimensions, and free from lamination, buckles, and surface, 
edge or other defects. It shall show in test pieces, for plates 
and shapes, an ultimate strength of not les3 than 58,000 lbs., 
nor more than 65,000 lbs., a yield point of 38,000 lbs., an 
elongation, for plates under 36 inches wide, of 26%, and for 
plates over 36 inches wide, of 24%, and a reduction of area 
of 50%. 

Rivet rod shall have aa ultimate strength of between 
50.000 and 54,000 lbs., with a reduction of area of 60%. 

46. Punching and Drifting Tests. — Punching test, ap- 
plied to steel plate, for stand pipe or boiler work, having 55,000 
to 65,000 lbs. tensile strength, a yield point of not less than 
30,000 lbs., and in ^ inch plate an elongation of 25% longitudin- 
ally, or 22% transversely: ^A piece 1% in. by 10 in. shall permit 
of punching a row of not less than eight ^ inch holes, ij^ inches 
centre to centre, without crack. Drifting test: -A piece 3 in. by 
not less than 5 in. shall undergo the punching of not less than two 
^ inch holes, 2 in. centre to centre, and i}4 inches from edges, 
said holes to be then enlarged cold by a sledge and drift-pin to at 
least i}( in. diameter, without cracking. The steel must stand 
cold hammering or scarfing at lap joints without cracking. It is 
hardly necessary to add, that it shall bend cold 180° on itself with- 
out crack. 
4 



34 STRUCTURAL MECHANICS. 

47. Effect of Shearing and Punching. — Steel, other tfian 
of the softest grade, is thought to be weakened by shearing and 
punching, by the development of minute cracks, which, how- 
ever, do not extend to any distance. To remove this weak- 
ened portion the edges of sheets are often planed after shear- 
ing, and punched holes are reamed. The specification for 
medium steel will read: — 

All sheared edges of plates and angles shall be planed off 
to a depth of one-fourth of an inch. All punched holes shall 
be reamed to a diameter one-eighth inch larger, so as to 
remove all the sheared surface of the metal. 

Sharp re-entrant corners are not allowed in good prac- 
tice. 

48. Stone. — Stone for building differs very much in com- 
position and quality. The igneous or primary rocks are gen- 
erally satisfactory in the matter of strength, but their hardness 
makes them expensive to fashion to other than simple forms. 
Stratified stones, like the limestones and sandstones, are found 
in all grades, from the hardest and most durable to the softest 
and most perishable. The only sure test of the ability of a 
building stone to resist climatic changes, to stand the weather, 
is the lapse of time. Artificial freezing and thawing of a small 
specimen, frequently repeated, will give indication as to dur- 
ability. 

Sound hard stones, like granite, gneiss, clay-slate, mar- 
ble, compact limestone, and the better grades of sandstone, 
are sufficiently strong to carry any loads brought upon them 
in ordinary buildings. In exceptional cases, special investi- 
gation should be made. 

Stratified stones should be laid on their natural bed, that 
is, so that the pressure shall come practically perpendicular to 
the layers. They are much stronger in such a position, and 
the moisture which porous stones absorb from the rain can 
readily dry out. If the stones are set on edge, the moisture 
is retained and, in the winter season, tends to dislodge frag- 
ments by the expansive force exerted when it freezes. Some 
sandstone facings rapidly deteriorate from this cause. Crys- 
tals of iron pyrites occur in some sandstones and unfit them 
for use in the face of walls. The discoloration resulting from 



MATERIALS. 35 

their oxidation, and the local breaking of the stone from the 
swelling are objectionable. 

49. Masonry. — Most masonry consists of regularly 
coursed stones on the face, with a backing of irregular shaped 
stones behind, Stones cut to regular form and laid in courses 
make ashlar masonry, if the stones are large and the courses 
continuous. When the stones are smaller, and the courses 
not entirely continuous, or sometimes quite irregular, although 
the faces are still rectangular, the descriptive name is somewhat 
uncertain, as block-in-course, random range, etc., down to 
coursed rubble, where the end joints of the stones are not per- 
pendicular to the beds. Rubble masonry denotes that class 
where the stones are of irregular shape, and fitted together with- 
out cutting. If the face of a stone is left as it comes from 
the quarry, the work is called quarry-faced or rock-faced. 
The kind of masonry depends upon the beds and joints. 
Walls of stone buildings have only a more or less thin 
facing of stone, the body of the wall being of brick. The 
stone facing should be well anchored to the brick work by 
iron straps. 

50. Specifications. — The following specifications will in- 
dicate how good work is described. Beginning with the por- 
tion under ground: — 

All foundation courses shall be built of rubble masonry, 
with selected, large flat stones not less than twelve inches 
thick, nor of less superficial surface than fifteen square feet. 
The foundation shall be brought to a level bed at the footing 
course, selected stones being used to give the six-inch 
offset required. The spaces between stones when laid close, 
shall at no place be over six inches in width, and these spaces 
shall be filled with small stones and spalls, flushed in cement 
mortar and well grouted. 

For rubble masonry above the foundation courses: — The 
stones used shall be quarried or split stones, hard, sound, and 
as nearly rectangular as possible, of good, flat beds and 
builds, and, unless used for trimming or closers, not less than 
six inches in thickness when laid on the largest face, and at 
right angles to the face of the wall. At least one-third of the 
stones shall be headers and extend back from the face not less 



36 STRUCTURAL MECHANICS. 

than three feet or clear through the wall. A proper alterna- 
tion of headers and stretchers shall be used in order to secure 
thorough bond. The stones shall be laid in full beds of 
cement mortar, with joints completely filled. The joints shall 
not exceed one inch in thickness, and the courses are to be 
properly leveled up. The stone shall be washed clean imme- 
diately before being laid. No dressing nor tooling will be 
allowed on any stone after it is in place. 

51. Ashlar Masonry. — The masonry above the footing 
course shall consist of quarry-faced ashlar, laid in horizontal 
courses, having parallel beds and vertical joints. The courses 
shall be not less than. 12 inches nor more than 30 inches in 
thickness, decreasing in thickness regularly from the bottom 
to the top of the wall, and shall be laid flush in cement mor- 
tar. Each course shall be thoroughly grouted before the suc- 
'<:eeding one is laid. The stones shall be alternately headers 
•and stretchers, and every header shall be laid immediately 
above a stretcher in the underlying course. iVll stones in 
every course shall bond at least 12 inches with those in the 
preceding course. The stretchers shall be not less than 3 
feet nor more than 6 feet in length, and not less than 2 
feet in width, nor less in width than one and a half times their 
depth. The headers shall be not less than 3! nor more than 
4j ft. in length, where the thickness of the wall will permit, 
and not less than ij ft. in width, nor less in width than they 
are in depth of course. 

Every stone shall be laid on its natural bed, which bed 
shall be well dressed to a plane surface and made as large as 
the stone will permit. The beds and sides of the stones 
shall be cut before being placed on the work, so that the mor- 
tar j )iats on the face shall not exceed one-half inch. No 
hammering on a stone shall be allowed after it is set; any in- 
equalities shall be pointed off. The vertical joints shall be 
cut to the same joint for a distance of not less than 12 inches 
from the face. All corners and batter lines shall be run with 
a chisel draft two inches wide on each face. 

The backing shall be of good-sized well-shaped stones, 
laid so as to break joints and thoroughly bond the work in all 
.directions, and leave no spaces over 6 inches in width. These 



MATERIALS. 37 

spaces shall be filled with small stones and spalls, flushed in 
cement mortar and well grouted. 

All bridge seats and tops of walls shall be finished with a 
dressed coping course, the joints in which shall not exceed 
one-fourth inch. Each stone shall be not less than 4 feet 
long, and all shall be 12 inches thick. The bridge seats shall 
have four, and the coping two ijx i J inch wrought iron dowels 
to each stone. 

No stone shall be shifted in its mortar bed by bars, and 
no movement of the stone laterally after being placed upon 
the wall will be permitted. Each stone shall be lowered to 
place dry, and shall then be raised, a bed of mortar spread to 
receive it, and the stone lowered to place. 

Every surface to which mortar is to be applied shall be 
freed from dust and dirt and thoroughly sprinkled just before 
the mortar is spread. 

Not more than three courses of the abutment shall be un- 
finished at a time, and the backing shall be carried up with 
the facing, but never in advance of it. Each course of 
masonry shall be grouted as laid with a mixture of two parts 
sand, one part cement and no more water than is necessary to 
give the required fluidity. The grout shall be worked into the 
vertical joints thoroughly with suitable flat iron blades, until 
all air is- expelled and the joints completely filled. 

All outside joints of the masonry shall be raked out to the 
depth of one inch and neatly pointed with a mortar of one 
part Portland cement and one part sand. 

All mortar used in the masonry shall consist of two parts 
by measure of sand and one part of American cement equal to 
the best quality of freshly burned Rosendale cement, to which 
shall be added only water enough to form a paste, stiff enough 
to handle with a trowel. The mortar shall be mixed in small 
quantities, as required for use, and shall be used before it has 
taken an initial set. The sand shall be sharp and clean, free 
from loam and clay. 

52. Brick Clay. — Clay may be roughly stated to be sili- 
cate of alumina. There are different grades of clay, from 
some of which china and porcelain are made, from others fire 
bricks, and from others common, or building bricks. The last 



38 STRUCTURAL MECHANICS. 

clays are the most easily fusible. Good brick clay, thoroughly 
burned, will yield hard, well shaped bricks. A too fusible 
clay will not allow sufficient burning, and hence the bricks 
will be comparatively soft. Lime in the clay lowers the fus- 
ing point, and the presence of lumps of lime in the bricks is a 
serious matter, as such lumps, when the bricks are wetted and 
laid in the wall, will slake, swell, and break the bricks. 

53. Bricks. — A good brick should be straight and sharp- 
edged, reasonably homogeneous when broken, dense and heavy. 
Two bricks struck together should give a ringing sound. 
Bricks which have a smooth exterior have been pressed after 
molding, and are more expensive. Some bricks, such as are 
used for paving, are rriade from ground shale. 

Soft, under-burned bricks are very porous, absorb much 
water, and cannot be used on the outside of a wall, especially 
near the ground line, for they soon disintegrate from freezing. 
Hard-burned bricks are very strong and satisfactory in any 
place; they can safely carry six or eight tons to the square 
foot. 

The red color of common bricks is an accidental charac- 
teristic, due to iron in the clay. Such bricks are redder, the 
harder they are burned, finally, in some cases, turning blue. 
The cream-colored bricks with no iron may be just as strong 
and are common in some sections. 

The builder usually lays the face of the wall with the best 
bricks, and the interior may be filled with the softer bricks, 
and even with bats, if permitted. The workman is not likely 
to fill the end joints with mortar unless an inspector insists on 
it. A shove or push joint is sometimes specified to cover this 
point. 

Bricks differ much in size in different parts of the country. 
The thickness of walls and the size of stone trimmings are 
to be adapted to the width and thickness of courses of brick. 
A course of headers is usually laid after from four to six 
courses of stretchers; but sometimes headers and stretchers 
alternate in every course. 

54. Lime. — Lime for use in ordinary masonry and 
brickwork is made by burning limestone, or calcium carbonate, 
and thus driving off by a high heat the carbon dioxide and 



MATERIALS. 39 

such water as the stone contains. There remains the quick- 
Hme of commerce, in lumps and powder. This quick-Hme 
has a great affinity for water and rapidly takes it up when 
offered, swelling greatly and falling apart, or slaking, into a 
fine, dr3\ white powder, with an evolution of much heat, due 
to the combination of the lime with the water. The use of 
more water produces a paste, and the addition of sand, which 
should be silicious, sharp in grain and clean, makes lime 
mortar. The sand is used, partly for economy, partly to 
diminish the tendency to crack when the mortar dries and 
hardens, and partly to increase the crushing strength. The 
proportion is usually 2 or 2^ parts by measure of sand to one 
of slaked lime in paste, or 5 to 6 parts of sand to one of 
unslaked lime. As lime tends to air-slake, it should be used 
when recently burned. 

Some limes slake rapidly and completely; other limes 
have lumps which slake slowly and should be allowed time to 
combine with the water. It is generally considered that lime 
mortar improves by standing, and that mortar intended for 
plastering should be made several days before it is used. 
Small unslaked fragments in the plaster will swell later and 
crack the finished surface. The lime paste is sometimes 
strained to remove such lumps. 

Lime mortar hardens by the drying out of part of the 
water which it contains, and by the slow absorption of carbon 
dioxide from the air. It thus passes back by degrees to a 
crystallized calcium carbonate surrounding the particles of 
sand. Dampness of the mortar is favorable to the attain- 
ment of this result, and the mortar in a brick wall which has 
been kept damp for some time, will harden better than where 
the wall is dry. Dry, porous bricks absorb rapidly, and 
almost completely, the water from the mortar, and reduce it 
to a powder or friable mass which Vv^ill not harden satisfac- 
torily. Hence bricks should be well wetted before they are 
laid. 

Lime mortar in the interior of a ver}^ thick wall may 
not harden for a long time, if at all, and hence should not 
be used in such a place. Slaked lime placed under water 
will not harden, as may be proved by experiment. In 



40 STRUCTURAL MECHANICS. 

both cases, such inaction is due to the exchision of the carbon 
dioxide. Lime mortar should never be used in wet founda- 
tions. 

Plaster for interior walls is lime mortar. Hair is added 
to the mortar for the first coat, so that the portion which is 
forced through the spaces between the laths and is clinched 2X 
the back may have sufficient tenacity to hold the plaster on 
the walls and ceiling. 

55. Natural Cement. — Natiiral cement is made by 
burning almost to vitrification a rock which contains lime, 
silica and alumina, that is, one which may be considered a 
mixture of a limestone and a clay rock. The carbon dioxide, 
moisture and water of crystallization are expelled by burning. 
The hard fragments must then be ground to powder, the finer 
the better. If the rock contains the several ingredients in 
proper proportion, upon the addition of water to the powder 
a reaction promptly takes place, and the double silicate of 
lime and alumina is formed, with a certain amount of water of 
crystallization. The fine grinding is necessary for a thorough 
mingling of the particles. The hardening which indicates the 
above reaction is called setting, and, in cements which con- 
tain the proper ingredients, begins in from a few minutes to 
half an hour. Those cements which contain an excess of 
lime set more slowly. The hardening, the tensile and com- 
pressive strength, increase rapidly at first, and at a decreasing 
rate for months. 

As access of air is not required for the setting of cement, 
the reaction taking place when water is added to the dry 
powder, cement mortar is used invariably under water and in 
wet places. It makes stronger work than lime mortar, and is 
generally used by engineers for stone masonry. Its greater 
cost than that of lime is due to the necessity of grinding the 
hard slag; while lime falls to powder when wet. The propor- 
tion of sand is i, 2, or 3 to one of cement, according to the 
strength desired, 2 to i being a common ratio for good work. 
The sand and cement are mixed dry and then wetted, in 
small quantities, to be used at once. 

The slower setting cement mortars are likely to show a 
greater strength, some months or years after use, than do the 



MATERIALS. 4 1 

quick setting" ones, which attain considerable strength very 
soon, but afterwards gain but little. 

56. Portland Cement. — If the statement made as to 
the composition of cement is correct, it should be possible 
to make a mixture of chalk, lime or marl and clay in proper 
proportions for cement, and the product ought to be more 
uniform in composition and characteristics than that from the 
natural rock. Such is the case; and the artificial cement, 
obtained by carefully mixing balls of clay and marl or lime, 
burning the balls nearly to vitrification and finely grinding, is 
known as Portland cement, an article superior to the natural 
cement, such as the Rosendale, Akron or Louisville brands. 
Much Portland cement is imported from Europe. 

The addition of brick-dust from well burned bricks to 
lime mortar will make the latter act somewhat like cement, or 
become hydraulic, as it is called. Volcanic earth has been 
used in the same way. 

57. Concrete. — Concrete is a mixture of cement mortar 
(cement and sand) with gravel and broken stone, the materials 
being so proportioned and thoroughly mixed that the gravel 
fills the spaces among the broken stone; the sand fills the 
spaces in the gravel; and the cement is rather more than 
sufficient to fill the interstices of the sand, coating all, and 
cementing the mass into a solid which possesses in time as 
much strength as many rocks. It is used in foundations, 
floors, walls, and for complete structures. The broken stone 
is usually required to be small enough to pass through a 2 in. 
or 2i in. ring. The stone is sometimes omitted. 

The concrete should not be made very soft and wet, but 
rather mealy, and should be deposited in pla.ce as soon as 
mixed, in layers from six to ten inches thick, and rammed till 
well consolidated, indicated by water slightly flushing to the 
surface. 

The proportions for mixing can be ascertained by filling a 
box or barrel with broken stone shaken down, and counting 
the buckets of water required to fill the spaces; then empty 
the barrel, put in the above number of buckets of gravel and 
count the buckets of water needed to fill the interstices of the 
gravel; repeat the operation with that number of buckets of 



42 STRUCTURAL MECHANICS. 

sand, and use an amount of cement a little more than sufficient 
to fill the spaces in the sand. If the gravel is sandy, screen 
it before using", in order to keep the proportions true. 

A common rule is, one part cement, two parts sand and 
five parts of broken stone or pebbles by measure. In another 
case, 5^ c. ft. cement, 7 c. ft. sand and 27 c. ft. broken stone 
made a cubic yard of concrete. A good and easily remem- 
bered rule is — one part cement, 2 parts sand, 3 parts coarse 
gravel and 4 parts broken stone. 

58. Cement Specifications. — The following specifica- 
tions for cements are reasonable, and the requirements are 
often exceeded: — 

Natural cement, when tested with a No. 50 sieve, shall not 
leave a residue of more than 4^ by weight; on a No. 100 sieve, 
not more than 25^; and on a No. 200 sieve, not more than 50^ 
by weight. 

A pat of the same, ^ in. thick, at 60° to 70° F., shall take 
an initial set in not less than 10 minutes, and shall set hard in not 
less that 30 minutes. 

Neat cement briquettes, one day in water after hard set, shall 
show a tensile strength of at least 75 lbs. per sq. inch. After one 
day in air and six in water, 150 lbs.; 28 days, 250 lbs. Mixture 
of one part cement, one standard crushed quartz, 7 day test, 75 
lbs. Briquettes from mixing box, two sand to one cement, 40 lbs. 

For Portland cement, the residue as above shall be not more 
than i^ on No. 50 sieve, 15^ on No. 100, and 40 f^ on No. 200. 
A pat of ^ in. thickness in a stiff paste shall require at least 30 
minutes for an initial set. 

Neat cement briquettes, 24 hours in water after hard set, 
shall develop 125 lbs. tensile strength per sq. inch. After one day 
in air and six in water, 350 lbs.; 28 days, 450 lbs. Cement mor- 
tar, with 3 parts standard crushed quartz to i cement, one day in 
air and six in water, shall show 125 lbs. Portland cement bri- 
quettes from the mixing board, two sand to one cement, one day 
in air and six in water, shall show 150 lbs. 

The three sieves shall be made of wire cloth. No. 25, No. 40 
and No. 40 wire respectively, Stubbs gauge, and having 2,500, 
10,000 and 40,000 meshes per sq. inch. 

The modulus of elasticity for concrete is about 700,000; 

for neat cement, about 3,000,000. The cohesion of iron and 

concrete is about 600 lbs. per sq. inch; of stone and cement 

about 15 lbs. per sq. inch. Brick masonry will fracture 

through the bricks rather than the joints if laid in thoroughly 

good cement mortar. 



MATERIALS. 43 

The modulus of rupture, or f for breaking weight of a 
beam, may be taken as about twice the tensile strength of the 
mortar used. 

59. Masonry Laid in Winter. — Civil engineers generally 
discontinue masonry construction as soon as freezing weather is 
likely to occur; but contractors in cities frequently carry up brick 
walls in lime mortar, although a temperature of o° F. may be 
experienced. In such a case the lime should be slaked with hot 
water, the bricks should be heated and laid in hot mortar. A 
thaw during erection is injurious to a wall built in winter, but con- 
tinuous freezing is not deemed harmful. A man can lay only 
about half as many bricks at such a time. 

Salt is often dissolved in the water with which cement mortar 
is mixed when it is to be used in freezing weather. Much salt will 
weaken the mortar. 

Natural cement concrete will disintegrate for a short distance 
below the surface, if exposed to a northern winter; but Portland 
cement is unimpaired by the action of frost, if well laid. 



CHAPTER III. 

BEAMS. 

60. Beams: Reactions. — A beam may be defined to be 
a piece of a structure, or the structure itself as a whole, sub- 
jected to transverse forces and bent by them. If the given 
forces do not act at right angles to the axis or centre line of 
the piece, their components in the direction of the axis cause 
tension or compression; to be found separately and provided 
for; the normal or transverse components alone produce the 
beam action or bending. 

As all trusses are skeleton beams, the same general prin- 
ciples apply to their analysis, and a careful study of beams 
will throw much light on truss action. 

Certain forces are usually given in amount and location 
on a beam or assumed. Such are the loads, concentrated at 
points or distributed over given distances, and due to the 
action of gravity; the pressure arising from wind, water or 
earth; or the action of other abutting pieces. 

It is necessary, in the first place, to satisfy the require- 
ments of equilibrium, that the sum of the transverse forces 
shall equal zero and that the sum of their moments about any 
point shall also equal zero. This result is accomplished by 
finding the magnitudes and direction of the forces required at 
certain given points, called the points of support, to produce 
equilibrium. The supporting forces, or reactions, exerted by 
the points of support against the beam, are two or more, 
except in the rare case where the beam is exactly balanced on 
one point of support. For cases where the reactions number 
more than two, see § 122. 

61. Beam supported at two points. Reactions. — The 
simplest and most generally applicable method for finding one 
of the two unknown reactions is to find the sum of the 
moments of the given forces about one of the points of sup- 
port, and to equate this sum with the moment of the other 



, BEAMS. 45 

reaction about the same point of support. Hence, divide the 
sum of the moments of the given external forces about one 
of the points of support by the distance between the two 
points of support, usually called the span, to find the reaction 
at the other point of support. The direction of this reaction 
is determined by the sign of its moment, as required for 
equilibrium. The amount of the other reaction is usually 
obtained by subtracting the one first found from the total 
given load. 



V/-SOO 

c- 



A 



^ : ^ r ^ D 



£i-4 4 B fc o Aai 



P.=5oo riy.3. 2oo=P'^ P,\--3^A ~/8y^--Pz , Fij.S. PJjSo 

C B 

Thus, in the three cases sketched, Pj — W t-^; Pg = W — Pj. 

Examples. — Fig. 3. If W = 500 lbs., A B = 30 ft., and 

500 . 18 
B C = 18 ft.; Pj = = 300 lbs., P2 = 500 — 300 = 200 

lbs. 

If W =: 750 lbs., A B = 20 ft. and A C = 5 ft., 

Pi = '""0^"" = 937>^ lbs., and Pg = 750 — 937>^ ^ — i87>^ lbs. 

If W = 150 lbs., A C = 20 ft. and A B =: 5 ft., 

P^ _ -^- • -^ _ ^^Q Vo'?,., and P2 = 150 — 750 = — 600 lbs. Note 

the magnitude of Pj and Pg as compared with W when the dis- 
tance between Pj and Pg is small. Such is often the case when 
the beam is built into a wall. 

Where the load is distributed at a known rate over a cer- 
tain length of the beam, the resultant load and the distance 
from its point of application to the point of support may be 
conveniently used. 



i^lg. 


4- 


750 


• 25 


20 


Fig. 


5- 


150 


• 25 



100 I zoo 150 






- KVWWS-W^WM 



ZOO „ &0 



I' 



9,-1320 Fi^.6. /^80=p2 F!=66s^ f^i3-7. :i'4%-^P2 



^,6 ,--i 




Example.— li A B = 40 ft., A D = 8 ft., D E = 16 ft., and 
the load on D E is 200 lbs. per ft., W = 3,200 lbs., and C B = 

24 ft. Therefore Pj = ^'^^^ ' — —= 1,920 lbs., and P^ = 3,200 

40 

— 1,920 = 1,280 lbs. 



46 



STRUCTURAL MECHANICS. 



raE 



D £ K f 



P,--200 
A 




JZ6S.6 



If several weights are given in position and magnitude, 
the same process for finding the reactions, or forces exerted by 
the points of support against the beam is appHcable. 

Exa7Jiples. — In Fig. 7, P^ = (100. 18-)- 200. i6-|- 150. 13 
+ 300 . II + 50 . 8 + 80 . o) ^ 16 = 665^^ lbs. P2 = 880 — 
665^ = 214 ^ lbs. The work can be checked by taking mo- 
ments about A to find P2, the moment 100 . 2 then being negative. 
If the depth of water against a bulkhead, Fig. 8, is 9 ft., and 
the distance between A and B, the points of support, is 6 ft., A 
being at the bottom, the unit water pressure at A will be 9 X 62.5 

= 562.5 lbs., which 
may be represented 
Fia.io> by A D, and at other 
points will vary with 
the depth below the 
surface, or as the or- 
dinates from E A to 
the inclined line E D. 
Hence the total press- 
ure E A, for a strip one foot in horizontal width, will be 562.5 X 9 
-^ 2 ^ 2,531 1^ lbs., and the resultant pressure will act at C, dis- 
tant ^ A E, or 3 ft. from A. P^ = 2,531^^ x 3 -^ 6 = 1,265.6 
lbs., and Pj — 2,531.2 — 1,265.6 ~ 1,265.6 lbs., a result 
that might have been anticipated, from the fact that the resultant 
pressure here passes midway between A and B. 

Let 1,000 lbs. be the weight of pulley and shaft attached by a 
hanger to the points D and E, Fig. 9. Let the beam A B == 10 
ft., A D = 4 ft., D E = 4 ft., E B = 2 ft.; and let C be 2 ft. 
away from the beam. As the beam is horizontal, P^ = 1,000 X 
4 -=- 10 ^ 400 lbs.; P2 = 1,000 — 400 = 600 lbs., and both act 
upwards. The 1,000 lbs. at C causes two vertical downward 
forces on the beam, each 500 lbs., at D and E. There is also 
compression of 500 lbs. in D E. 

When the beam is vertical. Fig. 10, by moments, as before, 
about B, P] = 1,000 . 2 -^- 10 = 200 lbs. at A acting to the left, 
being tension or a negative reaction. By moments about A, Pg 
= 1,000 . 2 -h 10 = 200 lbs. at B, acting to the right. Or Pj -f~ 
P2 := o; . •. Pj = — p2- By similar moments, the 1,000 lbs. at C 
causes two equal and opposite horizontal forces on the beam at D 
and E, of 500 lbs. each, that at D being tension on the connection, 
or acting towards the right, and that at E acting in the opposite 
direction. These two forces make a couple, balanced by the 
couple Pi Pg. The weight 1,000 lbs. multiplied by its arm 2 ft. is 
balanced by the opposing horizontal forces at D and E, 4 ft. apart. 
There remains a vertical force of 1,000 lbs. in A B, which may all 
be resisted by the point B, when the compression in D E — 500 
lbs. and in E B = 1,000 lbs.; or all by the point A, when the ten- 



BEAMS. 47 

sion in D E = 500 lbs. and in D A = 1,000 lbs.; or part may be 
resisted at A and the rest at B, the distribution being un- 
certain. This longitudinal force may be disregarded in dis- 
cussing the beam, as may the tension or compression in the 
hanger arms themselves. 

62. Bending Moments. — If an imaginary plane of sec- 
tion is passed through any point in a beam, the sum of the 
moments of all the external forces on one side of that section, 
taken about a point in the section, must be exactly equal and 
opposite to the sum of the moments of all the external forces 
on the other side of that section, taken about the same point. 
If not, the beam would revolve in the plane of the forces. The 
moment on the left side of the section tends to make that por- 
tion of the beam rotate in one direction about the point of 
section, and the equal moment on the right side of the section 
tends to make the right segment rotate in the opposite direc- 
tion. These two moments cause resistances in the interior of 
the beam at the section (which stresses will be discussed under 
resisting moment), with the result that the beam is bent to a 
slight degree. Either resultant moment on one side of a plane 
of section, about the section, is called the bending moment at 
that point, usually denoted by M, and is considered positive 
when it makes the beam concave on the upper side. Ordin- 
ary beams, supported at the ends and carrying loads, have 
positive bending moments. 

If upward reactions are positive, weights must be taken 
as negative and their sign regarded in writing moments. 

Examples. — Section at D, Fig. 3, 10 ft. from B. On the left 
of D, and about D, Pi(= 300) • 20 — 500 . 8 = 2,000 ft. lbs., 
positive bending moment at D. Or, about D, on the right side of 
the section; P2(= 200) . 10 = 2,000 ft. lbs., positive bending 
moment al D. Usually compute the simpler one. 

Section at A, Fig. 4, W . C A = — 750 . 5 = — 3,750 ft. lbs. 
negative bending moment at A, tending to make the beam convex 
on the upper side. At D, 10 ft. from B, M ~ — Pg . 10 = — 
187}^ . 10 = — 1,875 ^t. lbs., negative because Pg is negative. At 
A, taking moments on the right of and about A, M — — 1871^4 . 
20 = — 3,750 ft. lbs., as first obtained. This beam has negative 
bending moments at all points. 

In Fig. 5, M at D is — 150 . 10 = — 1,500 ft. lbs. It is evi- 
dent that the bending moments at all points between C and A can 



48 STRUCTURAL MECHANICS. 

be found without knowing the reactions. If this beam is built into 
a wall, the points of application of Pj and P^ are uncertain, as the 
pressures at A and B are distributed over more or less of the dis- 
tance that the beam is embedded. The max. M is at A, and is 
— 150 . 20 — — 3,000 ft. lbs. It is evident that the longer A B is, 
the smaller the reactions are, and hence the greater the security. 

In Fig. 6, the bending moment at C will be Pj . AC — - weight 
on D C . ^2 D C = 1,920 . 16 — 200 .8.4= 24,320 ft. lbs. 
At E, M = 1,280 . 16 = 20,480 ft. lbs. 

In Fig. 7, the bending moments at the several points of appli- 
cation of the weights, taking moments of all the external forces on 
the left of each section about the section, will be — 

At C, M = — 100 .0 = 0. 

At A, M = — 100 . 2 = —200 ft. lbs. 

At D, M = —100 . 5 + (665^ — 200) . 3 =: 896;^ ft. lbs. 

At E, M = — 100 ; 7 + 4659/8 • 5 — 150 . 2 = 1,328^ ft. lbs. 

At F, M = — 100 .10 4" 465^ . 8 — 150 . 5 — 300 . 3 = 
1,075 ^t. lbs. 

And, at B, M will be zero. M max. occurs at E. 

Do not assume that the maximum bending moment will 
be found at the point of application of the resultant of the 
load. The method for finding the point or points of maximum 
bending moment will be shown later. 

The moments on the right portion of the beam may be 
more easily found by taking moments on the right side of any 
section. Thus at F, M = (P^ — 80) . 8 = (214I — 80) . 8 = 
1,075 ft. lbs. Find the bending moment at the middle of E 
F. 1,201 T6 ft. lbs. 

In Fig. 8, the bending moment at section C of the piece A E 
may be found by considering the portion above C. As the unit 
pressure at C is 6 X 62)^ lbs. = 375 lbs. per sq. ft., M at C = 
(P, = 1,265.6) . 3 — (375 x6^2).6-^3::= 1,546.8 ft. lbs. 
At the section B, M = —(3 X 62^^ x 3 ^ 2 ) X i == —2811^ 
ft. lbs. 

In Fig. 9, as Pj = 400 lbs., P2 = 600 lbs., vertical forces at 
D and E are each 5,00 lbs.; M at D = 1,600 ft. lbs.; M at E = 
1,200 ft. lbs. 

In Fig 10, as Pj = — 200 lbs. = — P2, and the horizontal 
forces at D and E are ± 500 lbs.; M at D = — 800 ft. lbs.; M at 
E = -f- 400 ft. lbs. The beam will be concave on the left side at 
D and convex at E. The curvature must change between D and 
E, where M = o. Let this point be distant x from B. Then 
200 . X — 500 (x — 2) = o . • . jr =: 3^ ft. 

The curved piece A B, Fig. 11, with equal and opposite 
forces applied in the line connecting its ends, will experience 



BEAMS. 49 

a bending moment, at any point D, equal to P . C D, this 
ordinate being perpendicular to the chord. 

63. Shearing Forces. — In Fig. 3, of the 500 lbs. at C, 
300 lbs. goes to A and 200 lbs. to B. Any vertical section be- 
tween A and C must therefore have 300 lbs. acting vertically 
in it. On the left of such a section there will be 300 lbs. 
from P] acting upwards, and on the right of the same section 
there will be 300 lbs., coming from W, acting downwards. 
These two forces, acting in opposite directions on the two 
sides of the imaginary section, tend to cut the beam off, as 
would a pair of shears, and either of these two opposite 
forces is called the shearing force at the section, or simply 
the shear. When acting upwards on the left side of the 
section (and downwards on the right side), it is called posi- 
tive shear. When the reverse is the case the shear will be 
negative. 

Examples. — In Fig. 7, where a number of forces are applied 
to a beam, there must be found at any section between C and A a 
shear of — 100 lbs.; between A and D the shear will be — 100 -\- 
665^ — 200 = -\- 365^ lbs.; between D and E the shear will de- 
crease to 365^ — 150 = 215^ lbs.; on passing E the shear will 
change sign, being 215^ — 300 = — ^^Yd, lbs.; between F and B 
it will be — 84^ — 50 = — 134^ lbs.; and, on passing B, it be- 
comes zero, a check on the accuracy of the several calculations. 

In Fig. 8, the shear just above the support B = 3 x 62)^ X 3 
^ 2 = 281^ lbs.; just below the point B the shear is 281^ — 
1,265.6 = — 984.4 lbs.; and just above A it is 1,265.6 lbs. The 
signs used imply that the left side of A E corresponds to the upper 
side of an ordinary beam. As the shear is positive above A and 
negative below B, it changes sign at some intermediate point. 
Find that point. 

In Fig. 9, the shear anywhere between A and D is -|-4*^'^ 
lbs.; at all points between D and E it is 400 — 500 = — 100 lbs.; 
and between E and B is — 600 lbs. The shear changes sign at D. 

In Fig. 10, the shear on any horizontal plane of section be- 
tween B and E is — 200 lbs.; between E and D is — 200 -f 500 =: 
4- 300 lbs., and between D and A is +300 — 500 = — 200 lbs. 
The shear changes sign at both E and D. 

64. Summary. — To repeat: — The shearing force at any 
normal section of a beam may therefore be defined to be the 
algebraic sum of all the transverse forces on one side of the 
section. When this sum or resulting force acts upward on 

5 



50 STRUCTURAL MECHANICS. 

the left of the section, call it positive; when downward, 
negative. 

The bending moment at any right or normal section of a 
beam may be stated to be the algebraic sum of the moments 
of all the transverse forces on one side of the section, taken 
about any point in the section as axis. When this sum or 
resulting moment is right-handed or clock-wise on the left of 
and about the section, call it positive. A positive moment 
tends to make the beam concave on what is usually the upper 
side. 

By a proposition in mechanics, any force which acts at a 
given distance from a given point is equivalent to the same 
force at the point and. a moment made up of the force and the 
perpendicular from the point to the line of action of the force. 
Then, in Fig. 7, if a section plane is passed anywhere, as 
between D and E, the resultant force on the left, which is the 
algebraic sum of the given forces on the left of the section, 
is the shear at the section; and this resultant, multiplied by 
its arm or distance from the point in D E, giving a moment 
which is the algebraic sum of the moments of the several 
forces on the left of and about the point, is the bending moment 
at the section. 

It is also evident that the resulting action at any section 
is the sum of the several component actions; and hence that 
different loads may be discussed separately and their effects at 
any point added algebraically, if they can occur simultane- 
ously. Thus the shears and bending moments arising from 
-the weight of a beam itself may be determined, and to them 
may be added the shears and bending moments at the same 
points from other weights imposed on the beam. 

The numerous examples already given show that formulas 
are not needed for solving problems in beams, and the student 
will do well to accustom himself to using the data directly. 
Formulas, however, will now be derived, which will some- 
times be convenient for use, and from which may be deduced 
certain serviceable relationships. 

65. Reactions. — If, in Fig. 7, / = distance between 
points of support; a = known arm of W about the point 
where Pg acts, a being -}- when measured to the left of P2 and 



BEAMS. 51 

— when measured to the right, and upward forces being posi- 
tive; and 2 is the sign of summation, 

Pi / — S W « = o; or Pi = S W ^ -^ /. 
P2 + Pi — S W = o; or P, =3 2 W — P^. 

The same formulas will give the reactions for a beam built in 
at one end only, if the distance between Pj and Pg is known. 
The two reactions will then have opposite signs. 

For a distributed load, weight zu per unit of length of 
the beam, Y ^ I =■ f wda . a, between the limits over which ucr 
extends. 

The unit load w may vary, as in Fig. 8, in which case it 
must be so used in the formula; or it may be constant per 
unit of length, as in Fig. 6. In the latter case 

-t 1 = — J- between the given limits ot a. 

If w is constant, and covers /, 



Pi = 



wa'^ -]/ w I 



2/ 



1/ w I 

n 



as is easily seen from consideration of symmetry. 

66. Shear.— The shear F, positive when acting upwards, 
at any section distant x' from the left end of the beam, being 
the sum of the transverse or perpendicular forces on the left 
of the section, is, for a beam fixed at the right end onl}^, Fig. 5, 

— F^' = So W; or for a distributed load — F^- = / wdji;, 

the integration not extending beyond the length covered by 
the load. 

If the distance ;r' includes a point of support, as in the 
ordinary cases of beams supported at two points. Figs, 3 
and 6, 

F^- = Pi — 2o' W; or F^- = V, — j wdx, 

for distributed loads. But, for sections to the left of Pj, the 
first term disappears, reducing F to the corresponding expres- 
sions above. 



52 STRUCTURAL MECHANICS. 

The shearing force in sohd beams is not of much signifi- 
cance, on account of the amount of material which usually 
resists it; but in girders and trusses its determination is often 
necessary. It will also be required for locating the points of 
maximum bending moment, when simple inspection does not 
show them, § 68. 

67. Bending Moment. — The bending moment M, posi- 
tive when right handed on the left of any section, tending to 
make the beam concave on its upper side, will be, at a section 
distant x' from the left end, the sura of the moments of the 
forces on the left of the section, taken about a point in the 
section. For a beam fixed at the right end only, Fig. 5, 

— Mx' = 2o Wj^c; or — Mx' = / wxdx 

for a distributed load, x being the distance of W or zvdx from 
the section, and the integration not extending beyond the 
length covered by the load. 

Similarly, for beams supported at two points, as usually 
understood. Figs. 3 and 7, 

Mx' =: Pj ^ — 2o Wx; or Mx' = ^^x — / wxdx, 

for distributed loads, ;ir being, in all cases, the arm of P^, W 
or wdx about the section in question, and the integration cov- 
ering the loaded portion only. The unit load w may be con- 
stant or variable. 

If Pj is not to the left of the assumed section, drop the 
term P^ x, and obtain the preceding equations. 

68. Points of Maximum Bending Moment Occur 
Where the Shear Changes Sign. — It will now be seen by 
comparison, that F is always the first derivative of M, that is 

Fx = —7 — ^- Hence, according to the rule for determining 

maxima and minima, the bending moment is always a maxi- 
mum (or minimum) at the place where the shear is zero or 
changes in sign. This criterion is easily applied to locate the 
points of M maximum. Pass along the beam from the left (or 
right) until as much load is on the left (or right) of the sec- 
tion as will neutralize Pi (or Pg), and the point of M max. is 



BEAMS. 53 

found. Its value can then be computed. If the weight at a 
certain point is more than enough to reduce F to zero, F 
changes sign in passing that point, and hence M max. occurs 
there. 

For a beam fixed at one end only, F changes sign in pass- 
ing P], and hence M max. is found at the wall. 

Examples. — M max. occurs in Fig. 3, at C; in Fig. 4, at A; 
in Fig. 6, at 17.6 ft. from A; in Fig. 7, at A, and again at E; in 
Fig. 8, at B, and again at a distance x from E such that 62^ x . 
Yo, X ^ 1265^ . • . ^ = ^/4o.5 = 6.36 ft. ; in Fig. 9, at D; and in 
Fig. 10, at D and again at E. The bending moments which may 
not have been found at some of these points can now be computed. 

The reader who is familiar with graphics can draw the 
equilibrium or bending moment polygons or curves, and the 
shear diagrams, and notice the same relation in them."^ 

The unit load may also be considered as the derivative of 
the shear; F therefore has maximum (or minimum) values 
where the external forces change in sign. 

The origin of coordinates may be arbitrarily taken at any point 
in the length of the beam, and general expressions may be written. 
If — w is the unit load, either constant or variable, 

r 

Fx' = — - \ wdx = Fq — wx , if w is constant; 

/ wdx^ = Mo + Fq ^' — , if w is constant; 

where Fq and Mq are the constants of integration, the values of F 
and M at the origin, and the integration applies to the loaded por- 
tion. 

A general expression for the bending moment at any point of 
any beam will therefore have the form M^ = A -f- B.r -f" Qx^\ 
where A is the bending moment at the origin; Ba: is the sum of the 
moments of all the single forces, including Fq , from the origin to 
the point in question; and Cx^ is the sum of the similar moments 
of the distributed forces about the same point. This equation is 
useful in more complicated cases than those at present under con- 
sideration. 

69. Working Formulas. — There are a number of com- 
mon cases, for which the values of F and M may be derived 
for convenient reference. 

*See Greene's Graphics, Part II., Bridge Trusses. 



54 



STRUCTURAL MECHANICS. 




1 



V 



I. Beam fixed at one end only, and projecting a length // 
weight W at outer end. Fig. 12. Distance x measured 

I from wall. 

Fa; = W, and is constant. 
^ "~"^^^ M, = — W(/ — x), which in- 
'~^' ' (g) creases with the distance from the 

free end and has a maximum for ;r = o; or M max. ^- — W/. 

II. Same beam carrying a uniformly distributed load of 
wl. Origin at fixed end. Fig. 13. 

F^ = w(l — x); F max. = w/, 
at wall. 

M^ = — ^zv(/ — xY; M max. = "^ Fi_g./3. 

— iw/^ = — ^(wl)l at wall. In these two cases the load and 
the sheai^ each change sign at the wall by the addition of the 
upward reaction; hence F max. and M max. occur at the wall. 

III. Same beam, having both a uniform load and a single 
weight at the outer end. Add I. and II. 

F max. = W + wl. M max. = — (W + \wl)l. 

IV. Beam of span /, supported at ends, carrying a single 
weight W in middle. Fig. 14. Origin at left point of 

support. 
'"^^^''^^^---^^^'^^^^^ By symmetry, P^ = iW. 




<- 



-t 



n_g.i^. 



F. = JW on left, and = JW 
— W = — JW on right of middle. 

M, = V,x = ^Wx on left, and = ^Wx —W{x — J/) = 
JW(/ — x) on right. As F changes in sign at distance J/, 
M max. = JW/, at middle. 

V. Beam of span /, supported at 
ends, carrying a single weight W at 
known distance a from left end. 
Origin at left end„ * Fig. 15. 



<- a - 




F, = W 



Fv = W 



right. 





/ 




/- 


— 


a 




/ 




/ 




a 



W — P, = 






on the left of W; and P^ — W = — 



W^ 



on 



W — j-^ X on left of W; and w(" 



~i~{^l — •^) on the right of 



/ — a 



I 



X 



(^x 



--,) 



W, 



V,{l — x). 




BEAMS. 55 

a{l — a) 
M max. (where F changes sign) = W -, , at W. 

This value may be memorized as the weight into the 
product of the two segments divided by the span. 

VI. Beam of span /, supported at ends, and uniformly 
loaded with ivl. Fig. i6. Origin , 

at left. ^szzz^ss^z^^m 

By symmetry, Pj = Pg = j4w/. Fij/. 16. 

Fs =^ y2wl — wx = ivi^y^l — x^. 
F = o, when x = ^/, F max. = 

Mx = yzwix — y2,wx^ = ywx[i — x). 

Mx varies as the product of the two segments. 
M max. (when x = yi) — y^wP ■= y^iwiyi. 

VII. Case VI. can be added to IV. or V., for a uniform- 
ly loaded beam, carrying in addition a single weight. For 
the combination V. and VI., M max. will be found between W 
and the mid-span, at a distance from the middle of W*^ -^ wL 
It is often easier to calculate M max. directly than to use a 
formula. 

VIII. Beam of span /, supported at ends, and carrying 

a single moving load W. Origin at left. If, at any instant, 

W is distant x from the left end, the shear at a point distant 

(/ — x') 
a from tbe left will be W -j when x is greater than a^ 

Wx 
and — —7— when x is less than a. Each of these values is a 

maximum for x = a, or the positive and negative shears at 

any point are maximum when the weight reaches that point, 

W(/ — a) , W« 
and are then -r and — —i—. 

In the same way, W at x from the left will cause a 

WU — x) 
moment at a point distant a from the left of -^ a, when 

Wx , . , 

X is greater than a, and of ~j—^^ — ^) when x is less than 

Wa(l — a) . . . , ^ 

a. Hence M max. at ^ = ^—j -. when the weight 

rests on that point. The values of M max. at successive 
points vary as the product of the two segments, or as the 



5^ STRUCTURAL MECHANICS. 

ordinates to a parabola with a vertical axis and vertex at mid- 
span, with middle value of ^ W/. 

Example. — A 2,000 lb. wheel rolls across a 16 ft. span. M 
max. at middle = y^ . 2,000 . 16 = 8,000 ft. lbs. At 2 ft. each 
way from middle M max. = 125 . 6 . 10 = 7,500 ft. lbs.; at 4 
ft. = 125 .4.12 = 6,000 ft. lbs., and at 6 ft. = 125 . 2 . 14 =: 
3,500 ft. lbs. 

IX. Beam of span /, supported at the 
^ — cf — >^-z-Q-- -> ends and loaded uniformly over a dis- 



<^x-^- PC -> Pz tance / — a from right end. Total load 
^'^■''^ = w{l — a). Origin at left. Fig. 17. 

^w{l—ay ^ ,; , ^ / w(P — a') 
V, = y2 -^ ^; P^ =w{l—a) — V,= % -^ ^. 

Fx = -^ — -J , when ^ < (2; and — ^ — -. — w{x — a), 

when X y- a. 



o, when x ■=^ a -^ 



2/ ~ 2/ 



Ms = r^ :jc =: ~ X, when x <i a\ and 

ivi^l — of w(x — ay w{l — x){lx — a^) 

~ 2I 2 ~~ 2/ ' 

when X y- a. 

w{P — a''Y P ^ a" 
M max. = ^-^2 J at the pomt x = -^ — . 

X. Same beam, the load advancing continuously from 
the right. 

In this case M^ will increase as the load advances, until 
a = X, and then will continue to increase until a = o, when 
Mx =^ ^wx(l — x) as in case VI. ; and the absolute maximum 
M at middle = |w/l 

Indeed, since; a single load anywhere on such a beam 
produces positive bending moments at all points, a complete 
uniform load will give maximum moments. The same is true 
for equally spaced, equal weights. For several independent 
weights, the maximum bending moment at a given point will 
occur when the weights are brought as near that point as pos- 
sible. But when the weights differ in magnitude and are 
arbitrarily spaced, as in locomotive wheel weights on a bridge, 
a different solution is required. § 70. 



BEAMS. 57 

Fx will increase as the load advances, until a = x^ and 
will then decrease, since a is now less than x in the second 
value of Fx above. Hence the maximum positive shear at 
any point of this beam, from a continuous moving load, occurs 
when the load covers the right segment, of the two into which 

the point divides the beam. Then F^ max. = — —\ and 

the absolute max. shear is at the point of support, where 
Fo = }2wL 

If the figure is imagined turned end for end, it follows 
that the maximum negative shear, from a continuous moving 
load, occurs when the left segment is loaded. Then F^ max. 

= —^ with a minimum value of zero and an absolute 

2/ 

maximum shear at the point of support of — \wl. 

These values for F^ max., varying as the square of the 
distance from one or the other end, may be represented by 
ordinates to parabolas, drawn with the beam as tangent. 

As the shear from a single weight, on a beam supported 
at both ends, is positive on the left and negative on the right 
of the weight, it will again be seen that a full load on the 
right segment will give maximum positive shear at a given 
point; and full load on the left segment, maximum negative 
shear. 

XI. Case VL can be added to X, for combined steady 
and moving uniform loads, often termed dead (= w) and live 
(^= w'^ loads, giving maximum values of 

Fx = ^(/^ / — '^) -+- '^ — —1 foi' positive shears, 

2/ 

which will, however, become negative near the right point of 

support; and 

' 2 

Fx = w(^y2,i — •^) — for negative shears, 

2/ 

which will, however, be positive near the left abutment. At 

any one point these two values may be termed the maximum 

and minimum values of the shear for that point. 

Mx max. = y^ (w'-f 7v) X [I — x). 

Example. — A girder of 60 ft. span, weighing 300 lbs. per ft. 
and subjected to a uniform live load of 1,200 lbs. per ft. will ex- 



58 STRUCTURAL MECHANICS. 

perience a max. moment at mid-span of ^ i>5oo . 30 . 30 = 675,000 
ft. lbs. and a min. moment of 135,000 ft. lbs. At the quarter 
span the moments will be 506,250 ft. lbs. and 101,250 ft. lbs., or 
y^ the preceding. The maximum shear at the abutments is 1,500 
. 30 = 45,000 lbs. The shear at quarter span ranges from 

45^ 15^ 

300(30— 15) -f 1,200 Y^ = 24,750 lbs., to 4,500 — 1,200 -^ 

= 2,250 lbs., and at mid-span is i 9,000 lbs. 

XII. Beam of span /, supported at both ends, and carry- 
.- ^^ jj^g ^^^Q weights W, symmetrically placed 

A^ ^® — 7 ® ^ \ at a distance a from either end. Fig-. 18. 

Fia.lS. ^\ 

jw ^^ w4 By symmetry P, = Pg = W. 

^' 'z'^T 'T'^A^"^^ ' ^- = ^"^' for ^ < ^; =0 in middle por- 
F'lS-i^-' tion; = — Won right. 

Mx =: Wjr on left; = V^x — - W(jt: — a) = W^ in middle por- 
tion, and is constant, as is required where F = o. 

Case of cross-floorbeam for single track bridge; also 
trapezoidal truss, as indicated, when symmetrically loaded. 

XIII. There is no difficulty in determining the reactions, 
shears and bending moments of cantilever beams combined 
with a beam supported at both ends, producing a beam sup- 
ported at two points and overhanging at one or both ends. 
Fig. 19. 

Let left cantilever = ;;?/ ft., carrying VV at extremity; 
right cantilever = n/it., carr3dng W at its end; distance be- 
tween supports = /; total uniform load = iu{in -\- i + i^Y- 
With moments about Pg, Pi, = W (;;/ -f- i) — W';/ + \wl 

The shear at a distance x from the left, if jc < ml, is 
Fx = — (W -f- ivx.) Upon passing P^, F^ = Pi — (W -f- wx), 
with a probable change of sign. It may change in sign again 
in the length /, and again finally when Pg is added. 

Maximum negative bending moments are likely to be 
found at P^ and P2, with a maximum positive bending moment 
somewhere in /, if anywhere. 

M at Pi =: — (W -f y^wifil^jfil; M at Pg = — (W -f y^wnl^nl. 

-|- M max. where Jt: = — in — — 11 -\- — — ■ /. 

w IV 2 

W and W may arise from the weight of connected beams 
or girders. An inspection of the preceding values will show 



BEAMS. 59 

the effect of an increase of load on such parts. The unit load 
w may also change, and either in or n or both may be zero. 

70. Position of Wheel Concentrations for M max. at 
any given Point. — Where loads have definite magnitudes and spac- 
ings, as is the case with the wheel weights of a locomotive, the 
position of the load, on a beam ^__..j n /p \ 

or girder supported at both //'f^''^^^ (fe O^^^'^^^ 
ends, to give maximum bend- ' A dh^c Y^^ ^^ ^ ^ ^^>^ ^\ ^ 
ing moment at any given sec- ' ,- . ^ 

tion may be found as follows : — /^.-2 (/. 

Let the given section be C, at a distance a from the left abutment 
of a beam A B, of span /, Fig. 20. / 

Let Rj be the resultant of all loads on the left of C, and act- 
ing at a distance x^ from the left abutment; let R^ be the resultant 
of all loads to the right of C, and acting at a distance x.2 from the 
right abutment. The reaction Pj at A, due to Rg alone, will be 
Rg x.2 ~ /, and its moment about C will be Rg ax^ -^ I. Similarly, 
P2 at B due to Rj alone will be R^ x^ -h /, and its moment about C 
will be Ri(/ — a)x^ -h /. Hence for the whole load we have 

M at C = R. — ^ + Ri C^— ^)^i . 

If the entire system of loads is advanced a distance d to the 
left, 

M' at C = R, ^^^-^ + ^) + R, i^l-^^^^^-^i). 

and M'Js greater than M at C, if Rg « > Ri(^ — ^)' oi" i^ 
^— > -T; or by composition, if ^ ~\ ^ > — . If the op- 

posite is true M is greater than M'. 

But (Rj 4- R2) -^- / is the average load per running foot on 
the span, and R^ -f- <2 is the average load per foot to the left of C. 
Hence the rule: — 

If the average load per foot of span is greater than the aver- 
age load on the left segment a, the bending moment at C will be 
increased by moving the system of loads to the left, and vice versa. 
A panel may be conveniently used as the unit in applying this 
treatment to bridge trusses. 

Since, for maximum bending moment at any section, a load 
must be at that section, place a load W„ at the given point and 
compute the above inequality, first considering W„ as being just to 
the right and then just to the left of the section. If the inequality 
changes sign, the position with W„ at the section is one of M max. 
The value of M max. can then be computed as in § 62. If, how- 
ever, the inequality does not change sign, move the whole system 
until the next W comes to the section, and test the inequality 
again. 



6o STRUCTURAL MECHANICS. 

It sometimes happens that two or more different positions of 
the load will satisfy the condition just explained, and, to determine 
the absolute M max., each must be worked out numerically. 
When there are some W's much heavier than others, M max. is 
likely to occur under some one of them. When other loads are 
brought on at the right, or pass off at the left, they must not be 
overlooked. 

71. Position of Wheel Concentrations for Maximum 
Shear at any Given Point. — The shear at point C in the beam 
or girder, as the load comes on at the right end B, will increase 
until the first wheel Wj reaches C. When that wheel passes C, 
the shear at that point suddenly diminishes by W^, and then again 
gradually increases, until Wg reaches C. Let R be the sum of all 
loads on the span when Wj is at C, and x the distance from the 
centre of gravity of the- loads to the right point of support B. The 
shear at C will be P^ = R >:r -^ /. If the train moves to the left a 
distance h, the space between Wj and W2, so that W2 has just 
reached C, the shear at C will be R(:r -\- b) -^ I — W^, plus a small 
quantity /, which is the increase in Pj, due to any additional loads 
which may have come on the span during this advance of the train. 
The shear at C will therefore be increased by moving up Wg, if 
R <5 -^ / -|- / > Wj, or (as/ can often be neglected), if 

^ b ^ ..J R ^ W, 

R — > W^ or — - > _-L. 
/ lb 

Hence, move up the next load when the average load per foot 
on the span is greater than the load on the left divided by the dis- 
tance between W^ and Wg. 

RV R' W 

Similarly, W3 should be moved to C if > Wg or — > — ?, 

/ I c 

R' being the sum of the loads on the span when Wg is at C, and 

c the distance between W2 and Wg. 

It is not necessary to take account oi p, unless the two sides 

of the inequality are nearly equal. 

Example. — Span 60 ft., weights in units of 1,000 lbs. 

I 2 3 4 5 6 7 8 9 10 

Weights = 8 i5'i5 15 15 9 9 9 9 815 
Spacing = 8' 6' 4>^' 4>^' 7' 5' 6' 5' 8' 8' 

To apply test for M max.. at 15 ft. from left, load advancing 
from right. With Wg at quarter span, load on span = 104. If 

Wg is just to the right, — - > or — - > — ; if W, is just to 

60 15 4 I 

the left, ^-— !^ > ^; therefore move up Wg to right of quarter 
4 I 

112 'Z'X 

span. W](, now is on the span. > -^. Consider Wg to be 

4- I 



BEAMS. 6l 

just to the left; then < ^ , or the inequality changes with Wg. 

4 I 

Pi =(8. 59 + 15 . 51 +45 • 4o>^ + 36. 21 + 8 . 5) -^ 60 = 64.26. 
M max. =(64.26)15 — 8 . 14 — 15 . 6 ^ 761,900 ft. lbs. 

To test for F max. at same point. Put W^ at quarter span. 

Load on span ^95. — > — Move up W^; load on span now 

60 8 

104. 1^ < 11. Inequality changes. P^ = (8 . 53 -f- 15 . 45 
60 6 

+ 45 • 34>^ + 36 . 15) ^ 60 = 53.2. 

F max. = Pi — Wj = 53.2 — 8 == 45,200 lbs. 

When these locomotive wheel loads are distributed to the 
panel joints of a bridge truss through the longitudinal stringers, 
which span the panel distance between floorbeams, the above rule 
is modified. 

The load in the panel D E, being supported directly by the 
stringers, is by that means carried to the joints D and E. The 

amount thrown on D, when Wg is at E, will be, W/ -=- ^j if N = 

number of panels in the span; and, as the reaction is K[x -\- b) 

X A- b W ^ 
-^ /, the shear in the panel of the truss R is — ■ — — L N. 

/ / 

Substitute this value in place of the previous one, and obtain 

R — > Wi — , or — > W,. 

Hence, move up the next load, when the whole load on the 
truss divided by the whole number of panels is greater than the 
load in the panel. The locomotive Avill advance farther into suc- 
cessive panels as it advances from the right, to give maximum 
positive shear. As modern panels are long, the leading wheel is 
not likely to pass D before the proper position is found. 

72. Maximum Bending Moment on a Beam under 
Moving Loads. — When a beam or girder of uniform cross-sec- 
tion, such as a rolled I beam, supported at its ends, is subjected to 
a system of passing loads, such as an engine, heavy truck or trolley, 
it generally suffices to determine that position of the system of 
weights which causes the absolute maximum bending moment, the 
section where it is found, and its amount. 

In the previous figure, let C be that section. Let R = result- 
ant of all loads on the beam and d = its distance from B; 
R^ = resultant of the loads to the left of C. The reaction at left, 
P^ = R^ ~ /; and, since the bending moment at C is to be a max- 
imum, the shear at C must be zero, or 

R^ - R, = o. .-. A = ?2. 

I ' Id 



62 STRUCTURAL MECHANICS. 

But the position of loads must also satisfy the condition of 
§ 70, since there is to be maximum bending moment at C, and 

— i = .'. — i = — L, or a ^ a. 

a I ad 

The point of absolute maximum bending moment therefore is 
as far from one end as the centre of gravity of the whole load is 
from the other. This rule may be written — When the middle of 
the span bisects the distance between the centre of gravity of the 
whole load on the span and one of the wheels on either side of the 
centre of gravity, the desired moment is to be found under one of 
the two wheels. 

Example. — Beam of span 24 ft. Two wheels 6 ft. apart, one 
carrying 2,000 lbs. and one 4,000 lbs., pass across. Centre of 
gravity is 2,000 . 6 -^. 6,000 = 2 ft. from the heavier wheel. 
Then this wheel is to be placed i ft. from mid span. Reaction 
= 6,000 . II -f- 24 = 2,750 lbs. M max. = 2,750 . 11 = 30,250 
ft. lbs. 

For beams fixed at one end and supported at the other or fixed 
at both ends, and for continuous girders, see Chapter VII. 

73. Compound Beam. — When one beam upon which a 
load is imposed is supported by another, the bending moments 
and shears acting upon the latter can be found by algebraic 
subtraction of those resisted at corresponding sections of the 
former from the moments and shears of the system as a whole. 
That such must be the case will appear from the consideration 
that the bending moments and shears depend upon the weights 
and their positions on the span, and not upon the form of the 
girder or truss. 




Thus in Fig. 21, by XII., §69, the bending moment 
between A and E, for the combination is W^t, and between E 
and F is W ((^ + ^) ; the shear from A to E is W, and is zero 
on E F. The shear from C to E in the upper beam is W, 
and the moment at E is ^b. Therefore the shear in the 
lower beam from C to D is zero, and the moment on that 
portion must be constant and equal to W {a -\- b) — W<5 
= W<^. The weights on the upper beam are therefore trans- 



BEAMS. 63 

ferred to C and D for the moments and shears on the lower 
beam. 

Put the weights at C and D and the secondary supports 
at E and F, and examine the case. 

In the combination of Fig. 22, by case V., § 69, 

^, . . 17 W . C B 

1 lie moment at K = — — A F • 

A B ^ ^' 

The combined moment at D = " AD; 

The moment at C — " AC. 

The moment at D in the arm which carries W is 
— W . C D. Hence the moment at D in A B must be 

W y -— - A D + C D I, which will be greater than the moment 

on A B at C. This moment must also be equal to the reac- 
tion at B into B D. 

In Fig. 23, the external forces being equal and directly 
opposed, the resultant bending moment must be everywhere 
zero; hence the bending moments at corresponding sections 
of the spring beams must be equal and of opposite signs. 

Diagrams drawn below the figures show the same points 
clearly. 

For beams of two materials: see §§ 112, 113. 

74.' Total Tension Equals Total Compression. — If a 
beam, loaded in any manner, and in equilibrium under the 
moments caused by the external forces, is cut perpendicularly 
across by an imaginary plane of section, while the right- 
handed and left-handed bending moments already shown to 
exist, § 62, continue to act, it is evident that the left and 
right segments of the beam can only be restrained from re- 
volving about this section by the internal stresses exerted 
between the material particles contiguous to the section. 
These stresses must be of such signs, that is tensile and com- 
pressive; of such magnitude, provided the material does not 
give way; and so distributed over the cross-section, as to make 
a resisting moment just equal to the bending moment at the 
section. For the former is caused by the latter and balances it. 

Since the moment arms of these stresses lie in the per- 
pendicular plane of section, the components to be considered 



64 STRUCTURAL MECHANICS. 

now will be normal to the section. The tangential compo- 
nents are caused by and balance the external shear. 

As the external forces which tend to bend a beam are all 
transverse to it, and have no horizontal components, the 
internal stresses of tension and compression which are caused 
by the bending moment must be equal and opposite, as re- 
quired for a moment or couple, and hence the total normal 
internal tension on any section mnst equal the total normal 
compression. 

When any oblique or longitudinal external forces act on 
a beam, there is always found that resultant normal stress 
on any right section which is required to give equilibrium. I 

75. Distribution of Internal Stress on Any Cross- 
Section. — It may be convenient in the beginning to consider j 
one segment of the beam removed, and equilibrium to be \ 
assured between the external moment tending to rotate the 
remaining segment and the resisting moment developed in the 
beam at the section, as shown in Fig. 24. 

If two parallel lines near together are drawn on the side 
of a beam, perpendicularly to its length, before it is loaded, 1 

these lines, when the beam is loaded to any reasonable amount 
and bent by that loading, will still be straight, as far as can 
be observed from most careful examination; but they will now 
converge to a point known as the centre of curvature for 
that part of the beam. 

An assumption, then, that any and all right sections of 
the beam, \>Qxx\g plane before flexure, are still plane after the 
' flexitre of this beam, is reasonable. If the right sections be- 
came warped, that warping would apparently cause a cumula- 
tive endwise movement of the particles at successive sections, 
especially in a beam subjected to a constant maximum bend- 
ing moment over a considerable portion of its span; and such 
a movement and resulting distortion of the trace of the sec- 
tional plane ought therefore to become apparent to the eye. 
Such a warping can be perceived in shafts, other than cylin- 
drical, subjected to a twisting couple, but cannot be found in 
beams. 

The lines A C and B D just referred to will be found to be 
farther apart at the convex side of the beam, and nearer to- 



BEAMS. 



65 




ftC 



gether at the concave side than they first were; hence a line 
G H, lying somewhere between A B and C D, is unchanged in 
length. If, in Fig. 24, a line parallel to A C is drawn through 
H, the extremity of the fibre 
G H which has not changed 
in length, K L will represent ^ ] 
the shortening which I L has 
undergone in its reduction 
to I K, and N O will repre- 
sent the lengthening which 
M N has experienced, in 
stretching to M O. The 

lengthening or shortening of the fibres, whose length was 
originally G H =: ds, is directly proportional to the distance 
of the fibre from G H, the place of no change of length, and 
hence of no longitudinal or normal stress. 

The diagram, Fig. i, representing the elongation or 
shortening of a bar under increasing stresses, shows that, for 
stresses within the elastic limit, equal increments of length- 
ening and shortening are occasioned by equal increments of 
stress. If this beam has not been loaded so heavily as to 
produce a unit stress on any particle in excess of the elastic 
limit (and no working beam, one expected to last permanently, 
should be loaded to excess), the longitudinal unit stresses be- 
tween the particles will vary as the lengthening and shorten- 
ing of these fibres, that is, as the distance from the point of 
no stress. Hence, at any section, the direct stress is uni- 
formly varying, with a max- 
imum tension on the convex 
side and a maximum com- 
pression on the concave 
side. 

The stresses on different 
forms of cross-section A C are shown in Fig. 25. The 
total tension on the section is always equal to the total 
compression. 

76. Neutral Axis. — The arrows in Figs. 24, 25 may 
be taken to represent the unit stress at each point of the cross- 
section, varying as the distance from the plane of no stress. 




Fig. 25. 



66 STRUCTURAL MECHANICS. 

and constant in the direction z. To locate the point or plane 
of no stress or netctral axis for successive sections: — 

Let f^ and f^ be the unit stresses of compression and ten- 
sion between the particles at the extreme edge of any section, 
distant y^ and y^ from the point of no stress. It is plain that 
fc '• ft '-- Jc • yt from similar triangles, and that the unit stress 
/ at any point distant jj/ from the point of no stress will be 



P = -ry, or 7- y, or, in general -~ y, 



from a similar proportion. 

If zdy is the area of the strip on which the unit stress p 

is exerted, z being the co-ordinate at right angles to x and y, 

f 
the total force on zdy will be pzdy — — yzdy — cyzdy, where 

c is a constant, the unit stress at a unit distance. 

As the total normal tension on the section is to equal the 
total compression, or their sum is to be zero, § 74, the condi- 
.tion may be written 



\ pzdy = c \ yzdy 



O. 



Therefore the sum of the moments zdy . y of the strips zdy 
about the axis of z must balance or be zero. Then the axis 
of z or neutral axis must pass through the centre of gravity 
of a thin plate representing the section, and the neutral axis 
of any section lies in its plane, in a direction perpendicular to 
tlie plane of the' applied external forces. The axes of the 
successive cross-sections make up what is known as the 
neutral plane of the beam. Although there is no longitudinal 
or normal tension or compression at that line of the cross- 
section, it experiences shear, as will be shown later. 

77. Resisting Moment. — The law of the variation of 
stress on the cross-section and the location of the neutral axis 
have been established. The resisting moment is caused by 
and is equal to the bending moment. The moments of all 



BEAMS. 67 

the stresses about the neutral axis Z Z is, since p has the same 
sign as y, and the moments conspire, 

M =/(+/) zdy (+^) +/(—/) zdy (— y) = \ pyzdy 



As /" -^ J'l denotes the unit stress at either extreme fibre 

divided by its distance from the neutral axis, and p — ~y, 

^ yl 

M = /f \hdy=ll; and/ = M>ii. 

yj -y, y^ ^ ' 

where I represents y^j/^^^ about the axis Z Z, lying in the 
plane of the section, through the centre of gravity of the 
same, and perpendicular to the plane of the external forces 
applied to the beam. I is termed in mechanics the moment of 
inertia of a plane area, and is usually one of the principal 
moments of inertia of the area. The integral will be of the 
fourth power, involving the breadth and the cube of the depth. 

For moments of inertia of plane sections, see Chap. V. 

As moments of inertia for plane areas are of the fourth 
power, and can be represented by fibh^, where h is the ex- 
treme dimension parallel to y, and b to z, and as y^ niay be 
written m'h, the resisting moment can be represented, if n' x 
m — n, l3y 

M = -^ = nfbh\ 
n being a fraction. For a rectangular section this becomes 

M •--:=/. — -- yji =\fbh^; 

\2 ' 6 

and for a circular section 

M =/. "l^ ^ %d = — //^=— /Rl 
■^ 64 / 32 4 

Examples. — A timber beam 6 in. x 12 in., set on edge, with a 
safe unit stress of 800 lbs. will safely resist a bending moment 
amounting to 800 . 6 . 12^ -^ 6 = 115,200 in. lbs. 

A round shaft, 3 in. in diameter, if / = 12,000 lbs. will have 
a safe resisting moment of 12,000 . 22 . 3^ ^ 7 . 32 = 31,820 in. 
lbs. 

For sections other than a circle or square, either b or h 
is usually assumed and h or b then found. If the ratio k 



68 STRUCTURAL MECHANICS. 

-^ / is fixed by the desire to secure a certain degree of stiff- 
ness (see "Deflection of Beams," Chap. V.), the unknown 
quantity is b. 

Example. — A wooden beam 12 ft. span, carries 3,600 lbs. 
uniformly distributed. M = ^ W/ = ^ . 3,600 . 12 . 12 = 
64,800 in. lbs. If / = 1,000, E = 1,400,000, and the deflection v 

is not to exceed -g^iyth of the span, from ~ = — „ ^ is obtained 

/ 48 Ej'i 

I 5 . 1,000. 2 . 144 

;t— — s t'} •'• ^^ = 13 in. Then assuming /^ =14 

600 48 • 1,400,000 .A ^ b t 

in., a practicable size, 64,800 = — ^-- — d . 14^; and ^ = 2 in. 

Economy of material apparently calls for as large a 
value of A as possible; but the breadth d must be sufficient to 
give lateral stiffness to the beam, or it may fail by the buck- 
ling or sidewise flexure of the compression edge, between 
those points where it is stayed laterally. The effect of load- 
ing as a beam a thin board set on edge will make clear the 
tendency. 

When the plane of the applied forces does not pass 
through the axis of the beam, a twisting or torsional moment 
is added, which will be discussed in § 93. 

78. Limit of Application of M = f I -f- yj. — The expres- 
sion for the resisting moment at any section of a beam, caused 
by and always equal to the external bending moment at that 
section, is applicable only when the maximum unit stress / 
does not exceed the unit stress at the elastic limit of the 
material. If / exceeds that limit, a uniformly varying stress 
over the whole section is not found, and the neutral axis may 
not remain at the centre of gravity. Hence, also, the substi- 
tution of breaking weights, obtained by experiments on beams 
which fail, in a bending moment formula which is then 
equated with / 1 -^- fu results in values of /, the then so- 
called modulus of rupture, agreeing with neither the tensile 
nor the compressive strength of the material, and therefore of 
but limited value. This formula is correct for the purpose 
of design and construction; but its limitation should be kept 
in mind. 

79. The Smaller Value of f -^ yj to be Used. — Since 
from similar triangles /^ -^ y^ = /t -^ Jt » it is immaterial which 



BEAMS. 69 

ratio is used for M for a given cross-section. But, in design- 
ing a cross-section to resist a given moment, if }\ and y^ are 
not to be equal, another consideration has weight. A numer- 
ical example will bring out the distinction. 

A beam of 24 in. span is loaded at the middle with a 
weight of 500 lbs. M max. will be ^W/ — 500 . 6 = 3,000 
in. lbs. If the depth of the beam is 5 in., and its section is 
of such a form that the distance from its centre of gravity to 
the lower edge is 2 in., and to the upper edge is 3 in., while 
1 = 4, then 3,000 = i /t . 4 or i /c • 4- Hence the max. 
unit tension /J = 1,500 lbs. per sq. in., and the max. unit 
compression /"c = 2,250 lbs. per sq. in. But , if the material 
of the above beam must not be subjected to a unit stress 
greater than 2,000 lbs. per sq. in., that unit stress will be 
found on the compression side; for 2,000 lbs. per sq. in. on 
the tension side would be accompanied by 3,000 lbs. per sq. 
in. on the compression side; and a unit stress of 2,000 lbs. 
compression is only compatible in this case with 2,000 . | = 
i»333 lbs. unit stress tension. The beam will safely carry 
only a moment of 2,000 .4-^3 = 2,667 i^- lbs. 

Hence, when designing, with a maximum allowed value 
of/, and using a, form of section where j/^ and j^c differ, take 
that ratio of / -^ y^ which is the smaller. For a few mater- 
ials, where /J. and f^ may be taken as differing in magnitude, 
as perhaps in cast-iron, use that ratio f^ -^ y^ or f^ -^ y^ which 
gives the smaller value. As the elastic limit in tension and 
compression for a given material is usually the same, use in 
computations the larger value of y^. 

80. Inclined Beams. — A sloping beam is to be treated like 
a horizontal beam, so far as resisting stress produced by that com- 
ponent of the load which is normal to the beam is concerned. The 
component of the load which acts along the beam is to be consid- 
ered as producing a direct thrust along the beam if taken up at the 
lower end; or a direct tension, if taken up at the upper end, or as 
divided somewhat indeterminately, if resisted at both ends. If 
this longitudinal force is axial, the mean unit stress/' caused by it 
is to be added to the stress/" of the same kind from bending 
moment at the section where this sum/' +/" will be a maximum. 
This point can easily be found graphically. If the section of the 
piece is the unknown quantity, it will commonly suffice to use the 
value of max. M to determine an approximation to/", and to cor- 



70 STRUCTURAL MECHANICS. 

rect the section by the resulting value of/' +/" at the point 
where the sum is largest. 

If the direct force at the end or ends is not applied axially, 
its moment at any section may augment or diminish the bending 
moment of the normal components of the load. 

Cases of inclined beams, for a given load and inclination, are 
better solved directly than by the application of formulas. 

Example. A wooden rafter^ 15 ft. long, has a horizontal pro- 
jection of 12 ft., and a rise of 9 ft., and it carries a uniformly dis- 
tributed load of 1,500 lbs. The normal component of this load 
will be r, 200 lbs., the component along the roof 900 lbs. The 

max. bending moment, at the middle, will be -^ ^ 

6 ^ 8 

= 27,000 in. lbs. If the safe stress is 1,000 lbs., the section to 

^i,- ^1 1 J 1 1,000 bJi^ J J., . 

carry this moment should be — • — 27,000, or d/r = 162. 

6 

If /^ = 3, /^ z=: 8 in. — . If the mean thrust, at the middle of the 
rafter, is 1,250 lbs., the max. thrust, at the bottom end, will be 
1,700 lbs., and the min. thrust, at the top end, will be 800 lbs. 
The section of max. fibre stress will be a very little below the mid- 
dle. But, if the rafter is 3 in. X 8 in.,/" from bending moment 

will be il^^^—^ = 844 lbs. Also/' = '-^^ = 52 lbs. Hence 

/' -\- f" = 896 lbs., a satisfactory result, if the rafter is stayed 
laterally by the roof covering or otherwise. 

81. Curved Beams. — An originally curved beam, at any 
given cross-section made at right angles to its neutral axis, so 
far as the resisting stresses to bending moments are concerned, 
is in the same condition with an originally straight beam at a 
similar and equal cross-section to which the same bending 
moment is applied. Any definite thrust or tension at its tvvo 
ends adds a moment at each right section equal to the product 
of the force into the perpendicular ordinate from the chord to 
the centre of the section, and a force, parallel to the chord, 
which force can be resolved into one normal to the section and 
a shear. Compare Fig. 11. 

82. Movement of Neutral Axis, if Yield Point is Ex- 
ceeded. — If it is assumed that cross-sections of a beam still remain 
plane after the yield point is passed at the extreme fibres, the 
stretch and shortening of the fibres at any cross-section will con- 
tinue to vary with the distance from the neutral axis or plane. 
Suppose then that the elongation per unit of length of the outer 
tension fibre has attained an amount equal to O L, Fig. i. The 



BEAMS. 



71 



unit stress on that fibre will be L N. A fibre lying half way from 
that edge to the neutral axis will have a unit stress K M. The 
total tension on the cross-section must be the area O M N L, O L 
now being the distance from the neutral axis of the beam to the ten- 
sion edge. Since the total compression on the section must equal 
the total tension, an equal area O L' M' must be be cut off by L' 
M' and the compression curve. The neutral axis must then divide 
the given depth of the beam in the ratio of O L to O L', shifting in 
this case towards the compression side. Had the compression 
curve been below the tension curve, the neutral axis would have 
shifted towards the convex side of the beam. 

Since L N is less than L' M', the unit stress on the extreme 
fibre on the tension side is the less. Hence this displacement of 
the neutral axis favors the weaker side. If such action continued 
to the time of fracture, it would account for the fact that the appli- 
cation of the usual formula, /I -^- }\, to breaking' moments gives a 
value of/ which lies between the ultimate tensile and compressive 
strengths of the material. It must be borne in mind, however, 
that the compression portion of the section increases in breadth 
and the tension portion contracts, quite materially for ductile sub- 
stances, thus adding to the complication. A soft steel bar cannot 
be broken by flexure as a beam at a single test. 

A rectangular cross-section also tends to assume the section 
shown in Fig. 26. The compressed particles in the middle of the 
width can move up more readily than they can laterally, making 

the upper surface convex as well as wider, and 

the particles below at the edges, being drawn r" ~~ ~~~S 

or forced in, are crowded down, making the \ / 

lower surface concave as well as narrower. xj.:-^-^ ...^^^--^ 

Hence the position of the neutral axis is f^ig. E6. 

uncertain, after the yield point has been passed 
on either face; but it is probably moved towards the stronger side. 

83. Cross-Section of Equal Strength. — When a mater- 
ial will safely resist greater compression than tension, or the 
reverse, it is sometimes the custom to use such a form of cross- 
section that the centre of gravity lies nearer the weaker side. 
Cast-iron alone is properly used in sections of this sort. See 
Fig. 25, section at right. Wrought iron or steel sections are 
occasionally rolled or built up in a similar fashion, but the 
increase in width of the compression flange is then usually 
intended to increase its lateral stiffness. 

If/j = safe unit tensile stress, and/, = safe unit com- 
pressive stress, the centre of gravity of the section must be 
found at such point that )\ : y^ — fi '• fc > when the given safe 
stresses will occur simultaneously at the section. By com- 



72 STRUCTURAL MECHANICS. 

position, J/^ : j/^ '■ /i = /t '■ /c '■ /t ~^ fc j so that the centre of 
gravity should be distant from the bottom or top, 

y^ = /i . — Zl , or_)'c = /^ 



/t+/c /tH-/c 

Example. — If f^ = 3,000 lbs., and f^ = 9,000 lbs., jFt ^ 
/? . 3,000 -^ 12,000 =z i^ A. It a cast-iron 1 section is to be used, 
base 10 in., thickness throughout of i in., and height of web /i% 
then by moments around base, }({/i' -|- i) = 

\o -\- h 12 

I i2t; . . 161; ,-- 3,000.16^.2 o 

-\- — ± -|- 10 . 1 -|- 5 • 4 = — -' M = ^ ^ — — 82,500 

12 4 4.3 

in. lbs., the moment that the section will carry, 

84. Beam of Uniform Strength. — As has been shown in 
§ JJ , the resisting moment may be put into the form M = 
nfbh^ where 11 is a numerical factor depending on the form of 
cross-section. If, then, for a given load, bh^ be varied at suc- 
cessive cross-sections to correspond with the variation of the 
external bending moment, the unit stress on the extreme fibre 
will be constant; the beam will be equally strong at all sec- 
tions, except against shear; and there will be no waste of 
material for a given type of cross-section, provided material is 
not wasted in shaping. 

Suppose, for example, that a beam is to be supported at 
its ends, to carry W at the middle, and to be rectangular in 
cross-section. 

By § 69, IV., the bending moment at any point between 
one support and the middle is \ Wx. Equate this value with 
the resisting moment. J W^ = | /b/i^. To make / constant 
at all cross-sections, b/i^ must vary as x from each end to the 
middle. If k is constant, /; must vary as x, or the beam will 
be lozenge-shaped in plan and rectangular in elevation. If, 
on the other hand, b is constant, /i^ must vary as x, and the 
elevation will consist of two parabolas with vertices at the 
ends of the beam and axis horizontal, while the plan will be 
rectangular. 

The section need not be a rectangle. If the ratio of b to 
A is not fixed, the treatment will be like the above; but, if that 
ratio is fixed, as for a circular section, or other regular figure, 



BEAMS. 



73 



b = c/t, and k^ must vary as the external bending moment, or, 
in the case above, as x. The cross-section of the cast-iron 
beam in the example of the previous section may be varied in 
accordance with these principles. 

The following table gives the shape of beams of rectan- 
gular cross-section supported and loaded as stated. 





M. 


bh^ VA- 


/l2 CONSTANT, 


h CONSTANT, 




RIES AS 


b VARIES AS 


h^ VARIES AS 


Fixed at one end, 


— WX 


X 


X, triangular plan. 


X, parabolic elevation. 


W at other. 




Fig. 27. 


Fig. 28. 


Fixed at one end, 


—% WX2 


x^ 


X~, parabolic plan, 


X~, h varies as x, 


uniform load. 




Fig. 29. 


triangular elevation. 










Fig. 30. 


f 


W^~^X 


X 


X ) 


X < 1 


Sup't'dboth ends! 
W at a. 1 

I 


I 




(.triangular plan, 


(.parabolic elevation. 


7 a ~ X) 


l — x 


Z,_a,| Fig. 31. 


I- J Fi"-3- 


Sup't'dboth ends, 


WX{1 — X) 


x(l — x) 


Xil — x) parabolic plan. 


X(l — X), circular or ellip- 


uniform load. 


2 




Fig. 33- 


tical elevation. 
Fig. 34- 



When a beam supported at both ends carries a single 
moving load W, passing across the beam, the bending moment 
at the point x, where the load is at any instant, = Wx (/ — x) 
-^ /. Such a beam will therefore fall under the last class of 
the above table. 




w Fi^.28- 




1^(3. SI. 




F,j.2^. 




Frg.30. 





Fi^.J3. 




Beams which can be cast in form or built up may be made 
in the above outlines, if desired. Some common examples, 
such as brackets, girders of varying depth, working-beams, 
cranks, grate-bars, etc., are more or less close approximations 
to such forms. Enough material must also be found at any 



74 STRUCTURAL MECHANICS. 

section to resist the shear, as at the ends of beams supported 
at the ends, Figs. 31 to 34. 

Where a plate girder is used, (see Fig. 96), with a con- 
stant depth, the cross-section of the flanges, or their thick- 
ness when their breadth is constant, will theoretically and 
approximately follow the fourth column of the preceding table. 
If the flange section is to be constant or nearly so, the depth 
must vary in the same way, and not as in the fifth column. 

Roof and bridge trusses are beams of approximate uniform 
strength, for the different allowable unit stresses and for 
changing loads. The principles of this section have an influ- 
ence on the choice of outline for such trusses, and the shapes 
of moment diagrams suggest truss forms. 

85. Allowance for Weight of Beam. — If a beam is long 
and heavy, its own weight may cause a noticeable unit stress. 
While this weight is usually at first approximated to, or assumed, 
and then added to the given external load, the beam may be treated 
as follows: — 

Design the beam or girder for the given load W, and compute 
its weight B', and breadth b' . As W is at present all the load the 
beam ought to carry, the proportion exists 

Weight of beam _ B' 
Entire load W' ^ 

and the net external load the beam ought to carry will be given by 
the proportion. 

Entire load _ W' 
External load ~ W' — B'* 

As the load which a beam will carry varies with the breadth, 

and as it is desired to increase the net load from W' — B' to W', 

the breadth must be increased in this ratio, or the new breadth b 

will give 

b W' . V ^' 

OY b = b - 



^' • W' — B' W' — B' 

As the weight, the net load and the gross load are increased 

in the same ratio, the weight B of the final beam, and the gross 

load W will be 

W' W/ ^ 

B =: B' 1^^ ; W = 



W' — B'' W' — B'' 

W' and B' should have an approximately similar distribution. 
If different working unit stresses are allowed for B' and W', mul- 
tiply W' by the ratio of its unit stress to that of B'. 






BEAMS. 75 

86. Distribution of Shearing Stress in the Section of 
a Beam, Pin, Etc. — It will be proved, in § 182, that, at any 
point in a body under stress, the unit shear on a pair of planes 
at right angles must be equal. Whatever can be proved true 
in regard to the unit shear on a longitudinal plane at any 
point in a beam must therefore be true of the unit shear on a 
tra7isverse plane at the same point. 

Fig. 35 represents a portion of a beam bent under any 
load. The existence of shear on planes parallel to E F is 
shown by the tendency of the layers to slide by one another 
upon flexure. Let the cross-section of the beam be constant. 
If the bending moment at section H, a point close to G, dif- 
fers from that at G, there will be a shear on' the transverse 
section, because the shear is the first derivative of the bending 
moment, § 6"^. The direct stress, here compression, on the 
face H F of the solid H F E G, will 
differ from that on the face G E, since __ ^ 
the bending moments are different, and / "zi^l pznr- ? ; 

that difference will be balanced by a / ^/ ^^ .brr y \ 
longitudinal horizontal force, or shear, / "^ ~-^ -~^^-g-_^^-^x vl- p( 

on the plane F E, to oppose the ten- A"^" — ^— — -Li_ j 

dency to displacement. If this force Fis-^5. 

along the plane E F is divided by the 

area E F over which it is distributed, the longitudinal unit 
shear will be obtained. It follows from the first paragraph 
that the unit shear at the point E on the transverse section 
G A must be the same. It is also evident that the farther 
E F is taken from H G, the greater will be the difference 
between the total force on H F and that on G E, until the 
neutral axis is reached, and that the unit shear on the longi- 
tudinal plane E F must increase as E F approaches B, the 
neutral axis. The same thing is true, if the plane is supposed 
to lie at different distances from the edge A. Hence, at any 
transverse section A G, the unit shear on a longitudinal plane is 
most intense at the neutral axis; and therefore the unit shear on 
a transverse section A G is unequally distributed, being great- 
est at B, the neutral axis, and diminishing to zero at A and G. 
Pins and keys, and rivets which do not fit tightly in their 
holes, and hence are exposed to bending, have a maximum 



76 STRUCTURAL MECHANICS. 

unit shear at the centre of any cross-section, and this shear 
must therefore be greater than the mean value, and must 
determine the necessary section. 

To find the mathematical expression for the variation of 
shear on the plane A G: — 

O B D C is the trace of the neutral plane. B D = E F 
sensibly = dx. B E = j/, B G = jFi- Breadth of beam at 
any point = z, at neutral axis = z^ . Normal or direct unit 
stress at the point E on plane A G = /. Unit shear at E 
= q\ maximum, at B = ^q. M and F = bending moment 
and shearing force at section A G. 

By§ 77, /= -^ and/ = y. 
The total direct stress on plane G E is 

/ pzdy = — / yzdy. (j.) 

The difference between M at the section through B and 
M at the section through D ™ust be Yd.r, smce M = /f^., 

by § 6%. The horizontal force on E F is the excess of (i.) 

^ dx fy\ 
for G E over its value for H F or / yzdy. Divide by 

^ J y 

the area zdx of F E over which this horizontal force acts, to . 

find the unit shear. 

q^ — —- \ yzdy. Hence ^0 = 7 — / y^dy. 

^^ J V ^^o J 

Since the mean unit shear = F -^- S, the ratio of the 
maximum unit shear to the mean will be found by dividing 
^oby F -^ S. 

fyi 

/ yzdy 
Max. unit shear S O'l . Jo 

- yzdy = -J- , 



Mean unit shear Izr, / ^ ^' ^ 



where r = radius of gyration of the cross-section, and 
f yzdy is the moment of either the upper or lower part of the 



BEAMS. 77 

cross-section about the trace of the neutral plane. Hence the 
max. unit shear will be 

b I ydy 
Rectangle, ^ ^ — — ^ ^,V I = ~ ' or 50% greater, 

r^ -^/(R2 —y^)2ydy 2 . 2 R3 4 

^'''^'7 . — XR^-2R " R^3 ^" T' ''"^'^' ^'^^^^'' 
Thin ring, approximately =z 2 or 100 per cent, greater than the mean unit shear. 

For beams of variable cross-section I will not be constant; 
but the preceding" results are near enough the truth for prac- 
tical purposes. 

Example. — A 4 in. X 6 in. beam has at a certain section a 
shear of 2,400 lbs.; the max. unit shear on both the horizontal and 

2 zLOO "? 

vertical plane, at the middle of the depth, is — • -^ ^ icq lbs. 

24 2 

on the sq. in. 

As shearing resistance along the grain of timber is much 
less than the shearing resistance across the grain, wooden 
beams which fail by shearing fracture along the grain at or 
near the neutral axis, at that section where the external shear 
is greatest. As the unit shears on two planes at right angles 
through a given point are always equal, the shearing strength 
of timber across the grain cannot be availed of, since the 
piece will always shear along the grain. 

Example — A cylindrical bridge pin 3 in. diam., area 7.07 sq. 

in., has a shear of 50,000 lbs. The max. unit shear is — ^ . — 

7-07 3 

= 9,430 lbs. per sq. in. 

If the apparent allowable unit shear is reduced one quar- 
ter, as from 10,000 lbs. to 7,500 lbs., the same circular sec- 
tion for a pin will be obtained in designing, as if the maximum 
unit shear were considered. For a rectangular section the 
apparent allowable shear should be reduced one-third. 

87. Variation of Unit Shear. — The distribution of shear 
on three forms of cross section is indicated in Fig. 36, where 
the ordinates show the unit shear at corresponding points. 
For the rectangle the curve is a parabola, as the breadth of 



78 



STRUCTURAL MECHANICS. 



the section is a constant. For the circle, the parabola ordin- 
ates are divided by a varying breadth, altering the curve as 
below. The curve for the I shaped section will be made of 
three parabolas as shown, but the unit shear on the flanges 
will be given in due proportion by dividing the ordinates of the 
dotted parabolas by the ratio of width of flange to thickness 
of web, giving the full curve. The unit shear for the I section 



2o 





Fig. 36. 

is thus shown to be practically constant over the web, and to 
differ but little from the unit shear found by dividing the 
total shear at a section by the cross-section of the web, as is 
usually done in practice. 

Examples. — i. Three men carry a uniform timber 30 ft. long. 
One man holds one end of the timber; the other two support the 
beam on a handspike between them. Place the handspike so that 
each of the two shall carry ^ of the weight. 

2. Three sections of water pipe, each 12 ft. long, are leaded 
end to end. In lowering them into the trench, where shall the 
two slings be placed so that the joints will not be strained? 
Neglect the extra weight of socket. 

3. Wooden floor joists of 14 ft. span and spaced 12 in. from 
centre to centre are expected to carry a floor load of 80 lbs. per 
sq. ft. Vt f ■=. 900 lbs., what is a suitable size? 2 in. X 10 in. 

4. One of these joists comes at the side of an opening, 4 ft. 
by 6 ft., the load from the shorter joists, then 10 ft. long, being 
brought on this longer joist at 4 ft. from one end. How thick 
should this joist be? 4 in. 

5. A cylindrical water tank, radius 20 ft., is supported on I 
beams radiating from the centre. These beams are supported at 
one end under the centre of the tank and also on a circular girder 
of 15 ft. radius. They are spaced 3 ft. apart at their outer ex- 
tremities. If the load is 2,000 lbs. per sq. ft. of bottom of tank, 
find the max. -f~ and — M on a beam. 

-f- 29,600; — 68,750 ft. lbs. 

6. Find b and h for the strongest rectangular beam that can 
be sawed from a round log of diameter d. 



BEAMS. 79 

7. An opening 10 ft. wide, in a 16 in. brick wall, is spanned 
"by a beam supported at its ends. The max. load will be a triangle 
of brick 3 ft. high at the mid-span. If the brickwork weighs 112 
lbs. per c. ft., find M at the mid-span. 44,800 in. lbs. 

8. In the above problem, write an expression for M at any 
point, ii w =■ weight of unit volume of load, a =: height of load 
at middle, ^ = thickness of wall, / = span and x = distance of 
point from the support. 

9. A trolley weighing 2,000 lbs. runs across a beam 6 in. 
wide and of 20 ft. span. What will be the elevation of a beam of 
uniform strength, and what its depth at middle, if / = 800 lbs ? 

12 in. -f- 

10. A round steel pin is acted upon by two forces perpen- 
dicular to its axis, a thrust of 3,000 lbs. applied at 8 in. from the 
fixed end of the pin, and a pull of 2,000 lbs. applied 6 in. from 
the fixed end and making an angle of 60° with the direction of the 
first force. Find the size of the pin, if / = 8,000 lbs. 

M = 20,784 in. lbs. 

11. A beam of 20 ft. span carries two wheels 6 ft. apart 
longitudinally, and weighing 8,000 lbs. each. When they pass 
across the span, where and what is the max. M? 57,800 ft. lbs. 

12. A floor beam for a bridge spans the roadway a and pro- 
jects under each sidewalk 3. If dead load per foot is w, live load 
for roadway w\ for sidewalk w'^, write expressions for -)- M max. 
and — M max. 



CHAPTER IV. 

TORSION. 

88. Torsional Moment. — If a uniform cylindrical bar 
is twisted by applying equal and opposite couples or moments 
at two points of the axis, the planes of the couples being 
perpendicular to that axis, the particles on one side of a cross- 
section tend to rotate about the axis and past the particles on 
the other side of the section, thus developing a shearing stress 
that varies with the tendency to displacement of the particles, 
that is, directly as the distance of each particle from the cen- 
tre. The unit shear then is constant on any ring, and the 
shearing stresses thus set up at any section make up the 
resisting moment to the torsional moment of the applied 
couple. As all cross-sections are equal and the torsional 
moment is constant between the two points first referred to, 
each longitudinal fibre will take the form of a helix. 

89. Torsional Moment of a Cylinder. — If the unit 
shear at the circumference of the outer circle. Fig. 37, of 
radius r^ and diameter d is q^, the value at a distance r from 
the centre will be, by the above statement, q = q^r -^ i\. 
The total shearing force on the face of an infinitesimal particle 

whose lever arm is r, and area rdrdd, will be ■^r'^drde, and 

rx 

its moment about the centre will be — r^drde. Hence the 
resisting moment 

'W r'drdd = y^r. q,r,^ = "^^ = 0.196 q,d' for a 

cylinder. 

As the quantity to be integrated is r^ . rdrdo, or is the 
summation of the products of the areas rdrdd by the squares 
of their distances from the axis, this integral is the moment of 
inertia of the section about an axis through its centre and 



TORSION, 



8l 



perpendicular to its plane, known as the polar moment of 
inertia, or J. If y and z lie in the plane of the section, and 
X lies in the axis of the shaft, y" ;;i!r^ — f my^ -\- f vi.z\ and 
in general J = I^ -|- I^ for any form of cross-section; for a 
circle J = 21^ = \~T\- 

Hence the resisting moment against torsion may be 
written T = ^J -^ i\, which resembles in form the resisting 
moment against flexure, but differs in using the polar moment 
of inertia of the cross-section for the rectangular one, and in 
having q^, max. unit shear in place of/", max. unit tension or 
compression. 

go. Torsional Moment of a Square Shaft.— If a 
square bar is twisted and the shear is assumed to vary on the 




cross-section with the distance of the particles from the cen- 
tre, j = -f h\ r, = h i/i and 

T = -, — -j^rj • ~F — 0.236 qJi\ 

This assumption is not correct. The unit shear is actually 
the greatest at the middle of each side. For rectangular 
sections the preceding treatment would be seriously in error, 
but for a square section the error is not important. The last 
coefficient should be about 0.208. For shafts the cylindrical 
form is now almost universal. See § 92. 

The exact investigation for sections other than circles is very 
involved. See "Theory of the Elasticity of Solid Bodies," 
Clebsch: translated from the German into French, with Notes 
by de St. Venant and Flamant; and Report of Chief of Engineers, 
U. S A., for 1895, p. 3041, Part IV. 

Example. — A round shaft, 2^ in. diarn. carries a pulley of 

30 in. diam.; the difference of tension on the two parts of the 

16 . 15,000 . 7 . 2'^ 
belt is 1,000 lbs. Then T = 1,000 . 15; ^1 = —^ — -3 

= 4,887 lbs. per sq. in., if the torsional moment is entirely car- 
ried by the section of the shaft on one side of the pulley. 

7 



82 STRUCTURAL IMFXHANICS. 

91. The Twist of a Cylindrical Shaft. — If dd is the 
small angle at the centre that the radius revolves in passing longi- 
tudinally a distance dx, the distortion is 7\dO -^ dx, and, as the 
pitch of the helix is regular, dO -^ dx ^ ^ x. If q^ is the unit 
shear at the point whose radius is r^, and the modulus of elasticity 
for shear, by definition, § 10, C = ^j ^ distortion, 



i\dd 


qx 


do 





^1 


q.x 


2q,l 


dx 


- c^ 


dx ~~ 


X 


C;V 


" iZd 



where / is the distance between the points of application of the 
two couples. As 

T = ^ ^1^^ for a round shaft, q^ — ——^ and 







32 T/ 10 T/ 

I = 7^' nearly. 



-Cd' 

If, for a square shaft, T = 0.208 q^li'', 

T/ 9.6 T/ 

^ = 0.104 C/.* " ~QJF~^ ^"^'^>^- 

Example. — If in the preceding example the length of shaft 
subject to this twisting moment is 30 ft. = 360 in., and C — 

10 . 15,000 . 360 .2^ . 

0,000,000, 6 = 7 — o.ic;4. lo reduce this 

^ 9,000,000 . 5* ^^ 

180 
angle to degrees multiply by or 57.3, giving 6/ = 8° 50', nearly. 

92. St. Venant's Equations for Torsion. — When a 
torsional moment is applied to a body whose section is not a 
circle, the following equations have been given by M. St. 
Venant for the max. unit stress produced, which is found at 
points of the boundary nearest the centre : — 

^1 = 1 — for a rectangle whose shorter side is b and 

I 2 

whose moment of inertia is taken through the centre of gravity 

about an axis parallel to the longer side. 

T . 

^j = J — h for an ellipse whose least semi-diameter is b and 

I 
whose moment of inertia is taken about the greatest diameter. 

The torsional angle for unity of length 

T T T T 

— about 40 — — ^, for rectangle, and = 4 -^ ±^^ 

for ellipse; where J = polar moment of inertia about axis 
through the centre of gravity, and S = cross-section. 



TORSION. 



83 



93. Effect of Twisting on a Beam. — Combination of 
bending moment M and torsional moment T. Fig. 38. 

The normal unit stress on the cross-section of the extreme 
fibre, from the bending moment on the beam or shaft, is /", ten- 
sion on one edge, com- 



/ 



-i 



^ 



T. 





pression on the opposite 
edge. The unit shear 
on the cross-section of 
the same extreme fibre 
is ^1- Refer to § 197, 
and remember that, at 
that point, there is an 
equal unit shear on the 
plane at right angles to the 
cross-section; then there 
are given /"and ^j on the cross-sectional plane, and ^^ on the plane 
at right-angles, to find the direction and magnitude of the new- 
principal unit stress. 

A B = plane .of cross-section; O N = /, N R = ^j, laid off 
in succession. Then O R = resultant unit stress on A B. O B =: ^^ 
on second plane, revolved 90° to make the two planes and normals 
coincide. Draw B R connecting the extremities of the two 
stresses. As its middle point falls on the middle point M 
of O N, 



/, = O M + M R. 

■A = 
A - 



M R- = M N2 + N Rl .-. § 197, 
O M — M R, 



which will be opposite in kind to p^, since M R > O M. The 
direction of /^ is parallel to the line bisecting the angle N M R. 

Since, by § 77, M = / I^ -^ y^, where I^ = rectangular 
moment of inertia of cross-section, and, by § 89, T = ^j J -^ y-^, 
where J = polar moment of inertia of the same section, and since 
J = 2 Iz for a circle or a square, § 89, (i.) may be multipled by 
I2 -^ j'l, and transformed to 

M, = iM_ -f |/(JM^_ + iT^) = 4(M -f ^/{W + T^') ) (2.) 
where M = original bending moment at the section, T = original 
torsional moment, and M^ = equivalent resultant bending moment 
for which the beam or shaft should be designed, so that the unit 
stress shall not exceed/ when both M and T occur at the given 
section. 

Some authors multiply (i.) by J -.- y^, and produce 

T, = M -h |/(M2 + T^) 

as an equivalent torsional moment, to be used when T is much 
larger than M. 



I 



84 



STRUCTURAL MECHANICS. 



As the section on which the new principal unit stress acts is 
perpendicular to a line bisecting the angle N M R, shown by the 
broken line through O, this section is not quite the same as the 
original right section, and hence a small inaccuracy is involved 
in (2.). 

Examples. — i. If the pulley of the previous example weighs 
500 lbs. and is 18 in. from the hanger, on a free end of the shaft, 
and the unbalanced belt pull of 1,000 lbs. is horizontal, the re- 
sultant force will be 1,118 lbs., and a bending moment of 20,124 
in. lbs. will be felt at the hanger. Then Mj = 1(20,124 -|- 
-|/^ (20, 124'-^ -)- 15,000-)) = |-(2o, 124 -\- 25,100) = 22,612 in. lbs., 

22,612 .7.-^2.2" 

^ — = 14,735 lbs. 



,3 



which will cause a fibre stress of 

22 . 5' 

2. The wooden roller of a windlass is 4 ft. between bearings. 
What should be its diameter to safely lift 4,000 lbs. with a 2 in. 
rope and a crank at each end, both cranks being used and /being 
800 lbs.? 8>^ in. 

3. Design a shaft to transmit 500 horse power at 80 revolu- 
tions per min., if ^ = 9,000 lbs. * <^ ^ 6 in. 

4. How large a shaft will be required to resist a torsional 
-moment of 1,600 ft. lbs. if ^ = 7,500 lbs.? If the shaft is 75 ft. 
long and C ■=■ 11,200,000, what will be the angle of torsion? 

I in.; 45°. 



S 



CHAPTER V. 

MOMENTS OF INERTIA. 

94. Moments of Inertia. — Values of I for the more com- 
mon forms of cross-section S. Also of r" = I -^ S. 

I. Rectangle, height Ji, base b. Fig. 39. Axis through 
the centre of gravity and parallel to b. 

I^ = / fzdy = b fdy = — bf\ 

Jb/f bh^ _ bh^ _ bVi 

-\- . ly . 

24 24 12 ^ 12 



bh' 




/^2 




•r- bh = 




12 




12 



For an axis through the centre of gravit}^ and perpendicu- 
lar to the plane, 

J = I^, 4- I, = ^(^'^ + /.2), and r' ^ J- {P ^ h% 
■^ 12 12 

II. Triangle, height h, base b. Fig. 40. Axis as above 
and parallel to b. 



h:b=^h-y:z; , = I-(^ A - y) 



— §/i . /.—?/! 



L= / 'fzdy = l. ' [-^h-y^fdy = l-^—'^y' 



-Ik " J -U ^ 3 



A J w. A V 243 324 243 .^24 / 

b/i' bh h^ 



16 , 2 , I "A ,^ <^/-5 

4- A_y^ /I V 243 324 243 324 / -^6 



36 ■ 2 1. 



36 



STRUCTURAL MECHANICS. 



III. Isosceles triangle, about axis of symmetry. Fig. 
41. Height along axis Ji, base b. 



h : \ b ^ z : \ b — y\ z := h [i — ). 



V\ 



¥ 



L = 2 



/i (i _ ^)fdy = 2h r 



y_ 

2b 



.-,¥ 



2 h 



V 24 32 ^ 



hb^ 

"48" 



24 



The sum of II and III will be the polar moment J, about 
an axis through the centre of gravity and perpendicular to the 
plane. 

I 2 V -^ ±-' 






I I'h 
6 V3 







r,g.3^- FljAO. 



Fij. 4A 



n^'Ak. 



F/5. 43. 



IV. Circle, radius R, diameter d. Fig. 42. If ^ = 
angle between the axis of z and a radius drawn to the extremity 
of any element parallel to z, 

jV = R sin 6] 1 = R cos 6; ^' = R cos Odd. 



Iz = 4R* / sin' cos' do = — 2R* . ^\ I sin 4^ — ^1 



-R* 

4 



-d' 



= 1 ;: R^ - rr R' = i R' 



16 



■dK 



The polar moment of inertia J, may be easily written if 
r' = variable radius. 



J 



1'' . 2-7' d?' = 



-P4 



-R 



1 R' 



Since I^ -|- ly = J, and ly = I^ by symmetry, \^ = |- - R* as 
before. 






MOMENTS OF INERTIA. 87 

V. Ellipse. Diameters ^ and ^. Fig. 43. 

As the value of z in the ellipse is to that of z in the circle, 
as the respective horizontal diameters, or as b to d, and as 
the moment of the strip zdy varies as the breadth alone, the 
ellipse havino- horizontal diameter b, height d, gives 

_-d' b _ -bd\ 2 _ T , -bd _ d' 
^ 64 ' d 64 ^ ^ * 4 16 

I, , by analogy = ^. ] = '^ (d^ + y-); r"- = ^{d^ + f). 

VI. The moment of inertia of a hollow section, when 
the areas bounded respectively by the exterior and interior 
perimeters have a common axis through their centres of 
gravity, can be found by subtracting I for the latter from I 
for the former. Thus, 

Hollow rectangle, interior dimensions b' and h\ exterior 
b and h]l, = —ibW - bdi^''^. 

Hollow circle, interior radius R', exterior radius R; I^ = 
^;t(R^— R/^. 

The moment of inertia of a hollow ring of outside diame- 
ter d and inside diameter d\ the ratio of d' to d being n, may 
be written 



i{-) d''-, 



95. Moment of Inertia About a Parallel Axis. — To 

find the moment of inertia V of a plane area about an axis z 
parallel to the axis z^ through the centre of gravity and distant 
c from it. 

By definition V = / (y 4- cYzdy = f y^^zdy + 2c f yzdy 
-j- c^ f zdy. The first term of the second member is I^, the 
moment of inertia about the axis through the centre of 
gravity; the second term has for its integral the moment of 
the area about its centre of gravity, which moment is zero; 





I, = l^-.id^ '"') = l<^ «v- 




As the cross-section S — ^ - (</^ — ^/2) = i^ - (i . 


if 


= x_i^„andI.= |K-?^) = |(.^-I^) 



STRUCTURAL MECHANICS. 



and the integral in the third term is the given area S. Hence, 
r = I^ -I- r S = (r^ -f c') S = r"' S. 

Example. — \y for rectangle, axis parallel to b^ is -\%bh^. V 
about base = ^^.bJi" + \h^ . bh = Wr'; and r" = \h\ 

The reverse process is convenient for use. 

I^ :^ J _ ^^ S = (r'^ — c'^ S = r'^ S. 

As the value of I about an axis through the centre of gravity 
is the least of all I's about parallel axes, it can readily be seen 
whether c^ S is to be added or subtracted. 

It is frequently necessary to divide areas, such as T, I 
and built iron sections, and those of irregular outline, into 
parts whose moments of inertia are known, each about an 
axis through its own centre of gravity; then, to the sum of 
their several I's, add the sum of the products of each smaller 

A I B I I h 



-—-z—-hWu 



y,T^i-w 









,i(h-y). 



- -^ 

-z— 



area into the square of the distance from its axis to the par- 
allel axis through the centre of gravity of the whole. This 
rule is an expansion of the preceding one. 

I2 for the whole = 2 I -f- S r S. 
If the value of Ij about an axis distant c^ from the centre 
of gravity is known, and it is desired to find I2 about a par- 
allel axis distant C2 from the centre of gravity a combination 
of the two formulas 

Ij = I2 -|- c^^ S, and Ig = I^ -|- <r^^ S gives 
l, = l,^ (^/ - c,^) S. ^ 

Example. — I for a triangle about an axis through the vertex 
parallel to the base is easily obtained, since z : b — y : h. 

rh 



Therefore I about vertex = / — pVv — — - 



Then I about base = 



bll' /I 4^ ,,, bh _ bh' 

— .4- ( -^ ) h- ^ 

4 V9 97 2 12 

g6. Moments of Inertia, Continued. — VII. L, T, 

Channel or TT section. Fig. 44. Area = b/i — b'h' = A 
— A'. To find j/q = distance of centre of gravity from axis 



MOMEMTS OF INERTIA. 89 

though \Jl, parallel to b. Area bh will have no moment 
about Z'Z' . 

b'h'{\/i — U') A'(// — h') 



K't 


A7 


2 (A A.') 


" 2 S* 


h h') + 1 


J — - 


2 J _ 





-'° "~ ^/z - ^7?' - 2(A — A') 

bW b'h''' 
Then Iz - 77 — 77- + bhy\ ~ b'h'\_\{h 

I about edge A B or C D of L = I, + (A — A') {^\h ± y^ )\ 
By interchange of b and //, values of I about A C and B E are 
obtained. These values doubled will give ly ,, for the channel 
and T sections. 

Example. — h = 6in., A' = 5in., <^ 

13 sq. in. A' = 35 in. / = i in. I'o 

2 . 13 

Iz = 48(3 + 1-8) — 35 (2.1 + 3.4) = 37.9. 

Formulas for such cases are of little value. In actual com- 
putations follow the general rule. 

VIII. I2, for such symmetrical sections as shown in Fig. 
45, can be readily calculated by writing the value of I for the 
the exterior bounding rectangle and subtracting the 
I's for rectangles indicated by the dotted lines. 

97. Axes of Symmetry. — The following facts 
have useful applications. If a plane area has two 



Sin. 


J 


b' 


^=- 


7in., 


S 


= 


35 • 


I 


:^ 


I 


35 in 


. - 


— . 



■■J 



— z 



axes of symmetry not at right angles to each other, 
its moment of inertia is the same about all axes lying in it 
and passing through its centre of gravity. Examples — equi- 
lateral triangle, square, regular pentagon, hexagon, etc. I 
may be calculated, therefore, about that axis which gives the 
simplest relations. 

Example. — Hexagon, side <2. Axis through opposite vertices. 
Each half composed of one rectangle a ^/f . a, and two triangles of 
base \a and altitude a -\/\. I for rectangle about base = y^Ji^ = 

3^* ViiY'^ ^ ^^^ 0^^ triangle about base = \j)h^ — ^^ . J^^i/d)^ 
I, for hexagon = 2(i«Vf + Fg^VI) = ¥Vi- 
Area, S = 2a\/l + ^Vl = 3^VI- ''" = l^'- 



90 



STRUCTURAL MECHANICS. 



The polar moment of inertia, about the axis x, is equal 
to the sum of the two moments about jj/ and rj. As, in gen- 
eral, y and z may have any directions at right angles to one 
another, the sum of ly and I^ must always be a constant for a 
given area 

Two sections which have the same value for I^ do not have 
the same resisting moment unless j/j is also the same in both 
cases. 

98. Resisting Moment about an Oblique Axis. — When 
the plane of the external forces passes through the axis of the 
beam but is not parallel to either h or b, the maximum values 
of /"or the value of M max. can be found as follows: — Fig. 46. 
The section is A B E F; its centre of gravity is G; the 
plane of the applied forces and of flexure is N N; Y Y and Z Z 

are the usual rectangular axes, and 
the angle of axis N N with Y Y is ^. 
Let )\ and j/g denote the dis- 
tances of axis Z Z from the edges 
A B and E F respectively and z^ 
■^ and Z2 the similar distances of axis 
Y Y from B C and A F. 




bending moment of 



If M 
•^ the external torces at this section, 

the component in the plane of j/ will be M cos ^, and that 
in the plane of z will be M sin 0. The unit-stress on the 
layer A B from the former will be /' = M cos Oj/^ -^ I^ and 
on E F, = M cos Oj/^ -^ I^ . The stress on B C from the lat- 
ter component moment will be /" = M sin Oz^ -h ly , and on 
A F, = M sin 0^2 ^ ly . 

The points A, B, and F have stresses equal to the alge- 
braic sum /' ± f" or 



/= M ( 



(jFj or JF2) cos (sj or z^ sin 



± 



L 



')■ 



It is plain that the corners or points B and F have the 
maximum unit stresses in the above figures, as the sign of the 
second term in the above formula for these points will be -}-. 

/at B = M {-^-^ -|- -!-= I, compression, if M is -f . 



MOMENTS OF INERTIA. 9I 

^ ,^ /^Vn COS 6 , Zo sin 0\ . .. ,^ . 

/at F = M 1-^^^- h ^ ), tension, if M is +. 

/at A = M (?i^' - JA^j. 

If _jj z= j^2 or Sj = 02, the expression is simpler. 

Example, — 8 in. steel I beam, purlin on a roof that slopes at 
30° to horizon. Purlin normal to roof, load vertical, span 12^ 
ft., area carried 12^ X 8 ft., 30 lbs. per ft. Total load = 3,000 
lbs. M = 3,000 X 150 -^ 8 =: 56,250 in. lbs. Width of flange 
4^; Iz = 57-8; ly = 4-35- 

r . /'4 X 0.866 , 2.12 X 0.<\ ^ r c \ \ 

f = 56,250 ^ ^ + 1^ ^ = 56,250 (0.06 + 0.24) 

^ 57-8 4-35 ^ 

= 16,875 lbs., a value which is somewhat too large, especially 
if the assumed load is not a liberal estimate and if wind pressure 
is to be added. 

gg. Moments of Inertia for Thin Sections. — Values of 
I for rolled shapes may also be approximately obtained by the 
following method, and, if the values given in the manufac- 
turers' hand-books are not at hand, will prove 
serviceable. ^ 

The moment of inertia of a thin strip or 
rod, Fig. 47, of length L and thickness ^ about 'S-^^- 

an axis passing through one end of, and making an angle 6 
with it, is the sarne as if ^ tl^ were concentrated at the extreme 
end. Xet / = distance along strip to any particle. 

I = / Id/ . P sin2 e = yi /V sin'^ = ys t\.y\, 







or one-third the area multiplied by the square of the ordinate 
to the extreme end. 

This expression might be derived from I for a rectangle, 
taken about one base. 

If the rod is parallel to the axis, and at a distance j/j from 
it, I =: /L . y\, since all particles are equidistant from the 
axis. By the application of these two formulas the following 
values are obtained. 



Hollow rectangle, sides b and h\ b parallel to axis. 
I for two sides, b, = 2^/(^/^)' = ^z bth'\ 
I for 4 pieces, ^/^, = 4 . Yz . Yzht . y^h^ = yi ih^. 

I = y,bth' -\- yith' = (3^ + Z^)/. ye/i\ r'= ^^^ . — 



92 STRUCTURAL MECHANICS. 

II. Hollow square, side h\ axis parallel to side. 
I = 2,3 //v. r- = ^i h\ 

III. Do. do. axis diagonal, y — /i -j/^. 

1 = 4. ys/a . /i' . y2 = 2/2 wt. r'- = y^HK 

IV. Hollow triangle, base h, sides, each a, altitude" = Ji' 
-j- yb'. Distance of centre of gravity from base, }\ 

= ; — r. Axis is parallel to base. 

2a A;- b 

2 J 7 2 
I ~ yi atJi- — {2a 4- ^)A'/- -— yzatJi^ 



V 2a -4- bJ 



2a -\- b 



2 a -\- b. 

V. Hollow triangle, axis through vertex, perpendicular to b. 

b'' 

\ = 2 . y^ . yzbt . yb- + 2 . y^at . yb' = {2a + by . — . 

r~ = — . 

12 

VI. Hollow circle, radius R. Polar moment 

= 2 77 R/ . R' = 2 - K'L 1, = y ] = - RV. 
r^ = - R3/ ^ 2 7: R^ = i^ R'^ = y^' 

VII. Hollow hexagon, side a, axis through opposite vertices. 

I = {2at + 4 . yiatWut" == ^ a^t. j-^ = -^a\ 

2 12 

VIII. Cross of equal arms. Same as hollow square, II., III., if 
each arm = k. 

IX. Angle, unequal legs, /i and b, about an axis parallel to b. 
y from vertex or angle = ■,) y^h — v = 



2[b Ar Uy ^ - 2{b-\-h) 



I = dt . r- -I /zV -f /i/{y/i ^yj 

_ /- bh' Ji' b-/i' \ _ 4/^ + h hH 

~ y^ib + hf ^ V2^ 4(^ + hfY ~~ b ^ h ' 12" 



2 _ Ab -y h h' 



(b 4- //)■'' 12' 
If axis is parallel to //, transpose b and /i. 

X. Angle, about an axis through centre of gravity of each leg. 
Least value of I. 

bVi^ 

1 = Mb + Ji)yr-t. ;-2 = ,; , , j^^. 



I 



MOMENTS OF INERTIA. 93 

XL Angle, equal legs, //. Make b ^^ h in IX. Axis parallel to 
one leg. y' = \h. 

I = J_ /rV. r' = A /zl 
24 48 

XII. Do. Do. Axis through centre of gravity of each leg. 

1 2 24 

XIII. I Beam, web h, each flange b, axis perpendicular to web. 

I ^ (2//. i/r -h — JfM = ^i_+i /,V. 
12 12 



_ 6^' -[- // /r 



.v2 _ 



2 (^ -|- /^ 1 2 

XIV. Do. Do. Axis along web, /i. 

I = 4 . i . i^^ . F = ¥'^- ''" = ¥' -12^ + h). 

XV. Channel, web h^ each flange <^. Axis perpendicular to web. 
Same as XIII. 

XVI. Do. Do. Axis parallel to web. / 

y' from back = -- — , — j. 
2 -\- h 

I ^ 2 . Ut . \b'' + 2bt{^\b — y'Y + hty-' 

_ 2{b^hY^bh b't .,_2(b^hf^bh 

■" {2b + hy * Y" ^^' ~ (2^ + hy ' ^' ' 

XVII. Z bar, web h, each flange b, axis perpendicular to web. 
Same as XIII. See Plate III., XIV. 

XVIII. Do. Do. Axis along web h. 

\ --^ 2 . \bt . b^^ ^- IbH. r' = I ^ — 



2b + /i 

The sections are supposed to be very thin and the aver- 
age thickness is to be used, found by dividin<^ the area by the 
sum of the given lengths of lines. If 7-" alone is desired, / may 
be neglected. 

Examples. — 4 X 6 X f !-• Axis parallel to shorter leg. 

16 + 6 6^ 3 14.8 X 8 

I = . -= 14.8. r^ = — - == 3.95. ' 

10 12.8 10 X 3 

Axis parallel to longer leg. I = '~ . ' ^ -- 5.6. 

10 12.8 

10 in. I beam, 5 in. flange, area 9.7 sq. in. 2/.' -|- // — 20 . • . / 
30 4- 10 5 

r= o.S. I = ^^ . 100 . ™ = 167. 

^ 12 10 



94 



STRUCTURAL M FX'HANICS. 



XIX. Circular Arc, axis through centre parallel to chord. Fig. 48. 

Length of arc 2R^/. 

ds \ dx ^=^ 'K \ y. . • . yds -.- ^dx. 

I about A B = t/fd's -= R// iv/.r = R/ . area A B C D 

=: K\0 + sin cos (i)t, 

since area = 2R^/.^R-f-2.1R sin . R cos 0. 

f yds 
Distance of centre of gravity of arc from A B, _>-' 



K/dx R . chord C D R sin ^y 



fds 



fds 



arc C D 




I — R-V(^y + sin cos 0) — 2 ^Ot . y"^ 

-OS/ , • sin^ 
^Ho -f sm cos — 2 )/. 

. S = 2R^y^; If /^ = 2 R sin 0, 

sin cos sin- 



FisAS. 



= R^ f i 4- 



20 



0' 



) 



/r- 



/^ I cos 2 ~\ 

Vsir?7y ^ sin ^ ~ ?V" 

Axis through centre, perpendicular to chord. 



XX. Do. Do 

Fig. 49. 

I = t fy'^ds = ^t / ydx = R/" . area segment 
— R3 (^(t; — gjn cos ('y)/. S = 2R^y^. If ^ =: 2R sin 
sin (J cos /y~\ <^- /" sin 2 



R^ 



(4- 



2^y 



J 8 sin'-^ \ 



2O 



) 



100. Spacing of Channels. — To find the distance ^/ which 
should separate two channels, so that ;'^ may be the same about 
both rectangular axes. 

I St. When the flanges are turned out. Notation as in XVI. 
Axis parallel to web. For both channels, distance apart being d, 

I = 2ht . i^^ + 4 . W H- ¥P - 4- i • ¥"^ 



= 1 (/z + 2l?)dH -^ J (4^ 4- 6dyy'L 
Axis perpendicular to web, by XV, 
6b + // 



<.- 



V^ i^: 



I 



. /i't. 



12 



8/,3 



Solve for d. 



Equate these two values and transpose 

(/i + 2/?)d' + 4/;V = 1//^^ ^ 2/?/r — lb 
2d. When the flanges are turned in. 

I = 2/U . 1^'^ + 4 . J . yP^ _ 4 . J(l^„ ^)3/ 

= I (/z 4- 2b)d''/ + J(4/; —(id)bH. 
. • . (/^ + 2/7)^/2 — 4^V = W + 2/;//'' — 1^1 



MOMENTS OF INERTIA. 95 

Example. — 7 in. channel, 2 in. flanges; flanges turned out. 
wd'^ -\- 16/^ = 289; ^/ = \\ in. — . 

Flanges turned in. \\d- — \(id = 289; d -^ 6 in. — . 

Examples. — i. Find the moment of inertia of a trapezoid, 
bases a and b, height h, about one base. 

2. A 12 in. joist has two mortises cut through it, VI., Plate 
II., each 2 in. square, and 2 in. from edge of joist to edge of mor- 
tise. How much is that section of the joist weakened? sV or 26%. 

3. A bridge floor is made of plates rolled to half hexagon 
troughs, X., Plate III., 6 in. face, 5.2 in. deep, 12 in. opening, 
\ in. thick. Find the resisting moment of a section 18 in. wide. 

20.8/ 

4. If that floor is 14 ft. between trusses and carries two rails, 
5 ft. apart, each loaded with 2,000 lbs. per running foot, what will 
be the unit stress? 7, 790 lbs. 

5. Six thin rolled shapes, web a, make a hexagonal column^ 
radius a, with riveted outside flanges, each b in width. Prove that 

., _ a:^ + 4(^? + bf 

12 (c? 4" 2/^) 



CHAPTER VI. 



FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 



loi. Introduction. — As the stresses of tension and com- 
pression which make up the resisting moment at any section 
of a beam cause elongation and shortening of the respective 
longitudinal elements or layers on either side of the neutral 
plane, a curvature of the beam will result. This curvature 
will be found to depend upon the material used for the beam, 
upon the magnitude and distribution of the load, the span of 
the beam and manner of support, and upon the dimensions 
and form of cross-section. It is at times desirable to ascertain 
the amount of deflection, or perpendicular displacement from 
its original position, of any point, or of the most displaced 
point, of any given beam carrying a given load. 

o Further, the investigation of the 

forces and moments wnich act on beams 
supported in any other than the ways 
already discussed requires the use of 
equations that take account of the 
bending of the beams under these 
moments. There are too man}/ un- 
F'ij^.SO known quantities to admit of a solu- 

tion by the principles of statics alone. The required equa- 
tions involve expressions for the inclination or s/ope of the 
tangent to the curved neutral axis of the bent beam at any 
point, and its deflection, or perpendicular displacement, at 
any point from its original straight line, or from a given axis. 
102. Formula for Curvature. — If, through the points 
A and B, on the neutral axis of a bent beam. Fig. 50, and 
distant ds apart, normals C D and K G to the curve ot this 
neutral axis are drawn, the distance from A B to their inter- 
section will be the radius of curvature p for that portion of 
the curve. If through A a plane F H is passed parallel to 




DEFLECTION OF SIMPLE BEAMS. 97 

K G, the distance F C will be the elongation, or H D will be 
the shortening, from the unit stress f, of the extreme fibre 
which was ds long before flexure. Cross-sections plane 
before flexure are plane after flexure, § 75. 

/ 
A O = p', A C = _yi; C F = — ^i", § 10. From similar tri- 

f Ev 

angles A C F and O A B, /> : ds = y\ \ —-ds, or p = ~. As, by 

M ^ J 



77, M = fl ^ y,, 



I _ M 
~J ~ El' 



the reciprocal of the radius of curvature, called the curvature 
or the amount of bending at any one point. 

103. Slope and Deflection. — If the curve of the neutral 
axis is referred to rectangular co-ordinates, x being parallel to^ 
the original straight axis of the beam, and v being perpen- 
dicular to the same, the differential calculus gives for the 

radius of curvature, p = • For very slight curvature, 

such as is found in practical, safe beams, ds along the curve 
may be assumed equal to dx along the axis of x. Then 

I ^^z; M 



p dx-' E I ■ 
As M is a function of x, as has been seen already, the 

first definite integral, -— , will give the tangent of the inclina- 

dx 

tion or the slope of the tangent to the curve of the neutral 
axis at any point x, and the second integral will give v, the 
deflection, or perpendicular ordinate to the curve from the 
axis of X. 

While the following applications of the operations indi- 
cated in the last paragraph are examples only, the results in 
most of them will be serviceable for reference. 

The student must be careful, in solving problems of this 
class, to use a general value for M, and not M maximum. 
The origin of co-ordinates will be taken at a point of support, 
such a point being definitely located; x is measured hori- 
zontally, V vertically, and — v denotes deflection downwards. 

The greatest deflection for a given load is v max. The 
greatest allowable deflection for a fibre stress /"is v^. 



98 STRUCTURAL MECHANICS. 

If M -^- I is constant, the beam bends to the arc of a 
circle. This happens where M is constant and I is constant, 
or where I varies as M. 

Example. — The middle segment of a uniform beam under 
§ 69, case XII. If W — 2,000 lbs., a — ^ ft. = 60 in. I = 
^bh^ =:6.I2^-^-I2=: 864, and E = 1,400,000, 

I 120,000 I o • o ,. 

— z= ^-7 ^ -^ and p = 10,080 m. = 840 ft. 



— -~~..jy u liidA.. — • — — , 

--- X --^^-^^ 3 EI 

no. iz. A v, = J^. 

^ 3E7I 

104. Beam fixed at one end; single load at the other; 
origin at the wall; length := /; x measured horizontally, v 
vertically. 

Mx =-W(/-:.); _! = _-_ (/_^). Let -^= A. 



dx^ EI' '' EI 

dv 
dx 



Tan. inclination, or slope, at x = — = — ^/ {I — x) dx =^ 

—A(/x — ^x' + C). 



At the point where x = o, the slope is zero, and therefore 

C = o. 

v^ = —A/(/x —1 x^)dx = —A{^/x'-~^x' + C^) 
z; — o, when x = o, . • . C^ = o. 

For X ^ L tan /, or max. slope = — A(/^ — \ P\ = ^f^^\ 

^ ^ ' 2 E I 

1 W/^ 

and V max., or max. deflection = — Ai\P — -/^) = — — -r^-^- 

" 3 E I 

To determine the maximum allowable deflection of a 

given beam consistent with a safe unit stress in the extreme 

fibre at the section of maximum bending moment, substitute, 

in the expression for v max., the value of W in terms of f. 

Thus, by § 77, M max. = — W/ = — . .-. 
vJ — • • and v^ = — 



yr ' 3EI 3EJ', 



DEFLECTION OF SIMPLE BEAMS. 99 

Example. — If / = 60 in., ^ = 4 in., -^ = 8 in., W = 800 lbs., E 

4.8^ 512 , 800 . 60^ . 3 

^ 1,400,000; 1 = = ; max. slope = 



12 3 2 . 1,400,000. 512 

800 . 6o'^ .3 . 1 •/. /- 

■=. 0.006; V max. = ^= 0.24 m. ; and, it / 

3 . 1,400,000 . 512 

^^. ^ A a ^- ' 1,200 X 60^ 

:= 1,200 lbs., max. sate denection = 

3 X 1,400^000 X 4 

= 0.26 in.* 

It will be seen that, for a given weight, the maximum 
bending moment varies as the length /; the maximum slope 
varies as /^; and the maximum deflection as /^ The slope 
and deflection also vary inversely as I, or inversely as the 
breadth and the cube of the depth of the beam. The maxi- 
mum safe deflection, however, consistent with the working 
unit stress/", varies as /^ and inversely as j/j, or the depth of 
the beam. These relationships are true for other cases, as 
will be seen in what follows. 

The ease with which problems regarding deflection are 
solved depends greatly upon the point taken for the origin, as 
it influences the value of the constants of integration. 

M max. — —{wl) . i/ = —^wP. 

{wl)P fP 
V max. — — ^ \ . Vi = ~~^. -WL X i 

105. Beam fixed at one end; uniform load of zv per unit 
over the whole length /; origin at the wall. 

Let B = ^. 

Slope a.t X = ^- = — B/"(/^ — 2/x 4- x^) = 
ax ^ ' 

_ B {^Px — Ix' + \x^ 4- C). 

When x ^^ o, ^~ = o; . •. C = o. 
ax 

v^ = _B(|/V — llx"" + tV^* -f C). 
When a: = o, X = o; . • . C == o. 

■^Cancel factors before reducing. 




I IT 



lOO STRUCTURAL MECHANICS. 

For X ■= l, tan. max. slope — — B(/'^ — /^ 4- y*) = — \. .^. \ 
and V max. := — B(i/^ — \l' -f iV/') = — ^^^^^^ 



8EI 



Again, for maximum safe deflection, consistent with unit 
stress /"in the extreme fibre at the dangerous section, 

M max. = — {wl) },l = ^. 

2/1 , {wl)P fP 

. • . wl =^ -; and v^ - ■ — 



y,r ' 8 EI 4Ej'i 

io6. Combination of Uniform Load and Single Load 

at one end of a beam -fixed at the other end. Add the corres- 
ponding values of the two cases preceding. 

M max. - — [W/ + \{wl)l\, 
tan. max. slope = — -:— [-J W/^ -|- \ (^wl)P'\\ 

Note, in the expression for z' max., the relative deflections 
due to a load at the end and to the same load distributed 
along the beam; and compare with the respective maximum 
bending moments. 

Example. -^li the preceding beam weighs 50 lbs., the addi- 

50 . 60"* . 3 

tional deflection will be ^—- = 0.005 ii^-» too 

8 . 1,400,000 .512 

small a quantity to be of importance. In the majority of cases, the 

weight of the beam itself may be neglected, unless the span is long. 

M max. = ^{wiy. 

5 {wi)P 

107. Beam Supported at Both Ends; uniform load of w 
per unit over the whole length /; origin at left point of sup- 
port. By § 69, VI. 

d "V 7.V 

Mx = \w(l — x\x\ — -^ = — — - ilx — .T^) = B (Zx — x^). 
"^ ax- 2 E 1 ' ^ ' 

— - = B(|/jc^ — \x^ -(- C) =: tan. slope at x. 




DEFLECTION OF SIMPLE BEAMS. lOI 

To determine the constant of integration, we see from Fig. i6 

that — - ^ o, when x = \L Then 
ax 

o = i/3 — M' + C, or C = — tWI .-. 

~ = B(|/x^ — ix' — rV/n. 
ax - ' 

V^ = Y>{\lx^ — i^x' — rhPx + C). 

As z; = o, when jc = o; C = o, and disappears. 

tor :r = o, or Jt" = /, tan. max. slope = i -^^ — t^V-? 

24b I 

the opposite signs denoting opposite slopes at the two ends. 

^7^ w ( I' I'' l'-\ 

V max. (when x = fl) = — ^^-p I — - — I = 

^ ^ ' 2E I V48 192 247 

384* EI • 

For maximum safe deflection, consistent with a unit stress 
/"in the extreme fibre at the middle section, 

/I 1/ 7w 78/1, 5 /r- 

y, «' ^1/ 48 E^i 

Examples. — ^^^1 pine floor- joist, uniformly loaded, section 
2 X 12 inches, span 14 ft. = 168 in., has deflected £ in. at the 
middle. Is it safe? E ^^ 1,500,000. By the last formula, 

5 / -14' • 12- 

1 = -^/ -— F; /= 2,300 lbs. 

* 48 [ 500,000 .6 

and the beam is overloaded. 

What weight is it carrying ? By formula for v max. 

5 14^ . 12^ . 12 

f = -^ wl r- wl — 5,248 lbs. 

* 384 1,500,000 . 2 . 12 

M max. = iW/. 

— WP , //' 

V max. = -^^FTT- ^'1 = 12 :5^- 
48 E I Eji/j 

108. Beam Supported at Both Ends; single load W at 
middle of span /; origin at left point of support. 




d'-o _ W 
dx^ ~" 2 E I 



M, == 1 W,<; -— z^ -^ ^ = A:c 



I02 STRUCTURAL MECHANICS. 

This expression will apply only from jt = o to jc = J/; 
but, as the two halves of the deflection curve are symmetrical, 
the discussion of the left half will suffice. 

^ = A(1^2 _^ C ); ^ =r o, when x = l/;.-. C = — y\ 

dx ^^ ' dx ^ ^ 

v^ z= A(Jjc^ — ^Px -|- C); z/ = o, when x =: o; .• . C =z o. 

For jjc = o, tan, max. slope = ^-t^-f- The limit x ■= I \s 

loE 1 

not applicable. 

^ W (P P \ WP 

V max. (when x = 1/) = -^^{-^ - ^J = 



2EIV48 16-/ 48EI 

/I 
For max. safe deflection, since JW/ = — , as before, 

W = -^, and ^, = ^^' 



y^l i2Ey, 

Notice the numerical coefficients of v max. in §§ 104, 105, 

108, and 107. They are i, i-, — - and -. M max. varies as i, 

' ^ ^ ^ 48 8 . 48 

1 1 and i-W/. 

Exajnple. — A 10 in. steel I beam 

of 33 lbs. per ft. and I = 162, span, 

il_^__~(J;) X _^____^ 15 ft., := 180 in., carries in addition a 

° p uniform load of 767 lbs. per ft. of span 

_-. and 6,000 lbs. concentrated at the 

— =^- — '- ^^ middle. What will be its deflection 

and the max. unit fibre stress ? From 

§§ 107 and 108, 



00 . 15 6,ooo~\ 180 



V max. = I — - — + — — - I ~- = 0.35 m. 

V 304 48 729,000,000 . 162 

-^ c^ . T^^ , ,^Tr /. 162 / 800 .IS , 6,ooo\ ^ 

From § 69, IV. and VI., -^ = j — ^ + I 180; 

5 ^8 4 J 

f — 16,667 Ihs. 

109. Single Weight on Beam of Span 1, supported at 
both ends; W at a given distance a from the origin, which is 
at the left point of support. 



?.i 



DEFLECTION OF SIMPLE BEAMS. 



103 



As the load is eccentric, the curve of the beam is unsym- 
metrical, and equations must be written for each portion, x 
< a and x > a. 



ON LEFT OF WEIGHT. 

P, = W '^; M. = mp^,. 



dh 



w 



dv 

z'x = A(i/x3 — ^ax^ -\-Cx -\- C^O 
Vx =■ o, when x =^ o', 



(!•) 



ON RIGHT OF WEIGHT. 

P2 = — ; Mx = — (/ — X). 



d'^V 



w 






E 1/ 



— = A(«/x - |«x2 + CO- 



(2.) 



C^^ =0. ■z^x = O, when x = /; .-. C^'' = 

For equations to determine the constants C and C, use 

the value x = a; when it will be evident that — for the left 

dx 

dv 
segment must give the same value as does — for the right seg- 

dx 
ment; and v -dX a must be the same when obtained from the 
left column as when obtained from the right. Therefore, 
from (i.) and (2.), 

laH _ 1^3 _^ (^ ^ ^2^ _ x^3 _^ (.,. oj. c ^ c. _^ 1^2^^ ^ ^^ 
^ then gives i^V — i^* + C « + \aH = I^V — la'' + C'^ — 
^^/B _ C7, or C = —\a^ — \al\ Therefore C" = \aH, and C 
= \a^l — \a^ — lal'^. Substitute above. 



dv SN rl — a 

dx 

W 



^x 



6 2 6 3 J 



x=:o,tan.max.slope= t — 7- — I 

EI/V2 6 3 J 

_ W« (/— rt) (2/— «) 

~ ~ 67e1 • 



dv 
dx 

W 

^^=E 



l_f alx^ _axs _asx_al^x^asl\ 



6 E 1/ (^ ~ ^) t"^^ ~(^^ — ^) x]. 

_Wa(/— ^)(/4-a). 



X = /, tan. max slope: 



6/ EI 



As a is assumed to be less than i/, and the substitution 

of :^ = <^ in the value of —— gives a slope which is negative, 

dx 

the point of v max. will be found on the right of W, and for 

that value of x which makes — on the right zero. Hence 

dx 

alx — \ax^ = la^ -\- g-^/l 



X 



2/x 4- /2 =: /2 _ ^^2 _ 1/2 ^ ^p _ ^2^ 



/—x = ^l(p — a'); 
which is the distance from the right point of support. Sub- 



I04 STRUCTURAL MECHANICS. 

stitute in the expression for v^ on the right, to obtain the 
maximum deflection. 

It should be noticed that, when the weight is eccentric, 
the point of maximum deflection is found between the weight 
and the mid-span, and not at the point of maximum bending 
moment, which latter is under the weight. 

no. Two Equal Weights on Beam of Span 1, sup- 
ported at the ends; each W, S3'mmetrically placed, distant a 
from one end. Fig. i8. 

This case may be solved by itself, but can be more readily 
treated by reference to § 109. Thus the maximum deflection 
will be at the middle, and can be found by making ^ = J/ in 
the above value of v for the right segment and doubling the 
result. Then 

The deflection under a weight will be given by the addi- 
tion of V dX a and v at (/ — a) of the preceding case. Thus 

^ at W = — -^^(3/— 4^). 

Example. — A round iron bar, 12 ft. long, and 2 in. diam., 

carries two weights of 200 lbs. each at points 3 ft. distant from 

either of the two supported ends. The deflection at a weight = 

200 . 36^ . 7 . 4 

— (3.12'^ — 4 • 3 • 12) = 0.56 m. the maxi- 

6 . 28,000,000 . 22 

.... . 200 .3.12.7.4 ^ .. 

mum unit bendmg stress is ^ = 9,160 lbs. 

22 

DEFLECTION OF BEAMS - OF UNIFORM STRENGTH. 

It will be apparent that a beam of uniform strength will 
not be so stifl as a corresponding beam of uniform section 
sufficient to carry safely the maximum bending moment; for 
the stiffness arising from the additional material in the second 
case is lost. 



DEFLECTION OF SIMPLE BEAMS. 



105 



III. Uniform Strength and Uniform Depth. — Since 
M = nfbh^ and varies as bh^, and I = n'bh^ and varies as bh^\ 
M -^- I varies as i -^- h. But if h is constant, M ^ I is con- 

stant and — - is constant. Therefore all beams of this class 
dx' 

bend to the arc of a circle. 

I. Beam fixed at one end only, and loaded with W at the 

other. § 84, Fig. 27. 

d'^v _ M _ ^V^ 

;^ ~ El ^ ~~ El (^~" ^^' 

If, in all cases, Iq = moment of inertia at the largest section, 
which is in this case at the wall, I at the distance x from the 

/ — X 



wall will be to I^ as / 
d'v W/ 



X 



is to /, or I = I 



/ 



Therefore 



, a constant, as stated above. 



dx"- E I^ 

Note that the quantity divided by E I^ is M max. 



dv 
dx 



_ W/ p 

~ eX./ o 



dx ^ — 



V max. = 



o 

W/ 



W/ 

et; 






E I 



t--v Ci/i^ — 



2E L 



a deflection 50% in excess of that of the corresponding 
uniform beam, while the 
max. slope is twice as great. 

Examples . — If a triangu- 
lar sheet of metal, like the 
dotted triangle in Fig. 51, is 
cut into strips, as represented 
by the dotted lines, and these 
strips are superimposed as 
shown above, the strips, if 
fastened at the ends, and sub- 
jected to W as shown, will tend 
to bend in arcs of circles, and 
will remain approximately in 
contact. If / = 10 in., <^ = 4 in., /? — ^ in., W = 400 lbs. and E = 

o 1 J £1 • -11 1 400 • 10^ . 4^ . 12 

28,000,000, the deflection will be = 1.37 m. 

2 . 28,000,000 . 4 

An elliptical steel spring 2 ft. long, of 4 layers as shown, 

each 2 in. broad and y^ in. thick, under a load of 100 lbs. at its 




'g 



51 



Io6 STRUCTURAL MECHANICS. 

•in -11 •,• T- in IO° • '^^^ • 8^- 12 

miuale, will, if b = 20,000,000, denect = 2.2. 

2 . 29,000,000 . 8 

rr^i • ,-1 -n 1 50 • 12 . 6 . 8^ ^ ^ 

in. The max, unit fibre stress will be = 28,800 

8 

lbs. Note that one-half of the weight is found at each hinge, and 

that the deflection of one arm is doubled by the use of two springs 

as shown. 

II. Beam fixed at one end only and uniformly loaded 
with za per unit. § 84, Fig 29. 

S = -^j(^--r- I:Io=^:^o=(/-.r:/^ 



dx 2E Iq / ^ 2E Iq 

wp n wP 

V max. = — T^ T / xdx = 



2EI07 ,^"' 4E lo' 

a deflection twice that of the corresponding uniform beam. 

In these two cases there are no constants of integration,. 

since — - and z/ = o, when x = o. 
ax 

III. Beam supported at both ends and carrying W at 
middle. § 84, Fig. 31, 

d'v _y^x J J _ 1. ^^ _ w/ 

— - = o, when jc = i/ ; . •. C = — i/. 

dx ^ ^ 

W/ , W/' 

Z' = — ^T-r {\^^ — i/jc). When jt: = 1/, z; max. = ^^ -, ? 

4EIo^^ ^ ^ ^ 32E lo 

a deflection 50% greater than for a corresponding uniform 
beam. 

IV. Beam supported at both ends, and uniformly loaded 
with w per unit of length. § 84, Fig. 33. 

g = j^ {Ix-x^) ; I : lo = X (/ - ^) : \l- ; 

ax orLr Iq 

-- = o, when x = 4/ :. •. C = — i/. 
dx ^ ^ 

wP '^^P 

V = gE-r^*"'' ~ 4"^^^- ^^^"^ -^ = i^, ^^ max. = — ^^^ j^ r 



DEFLECTION OF SIMPLE BEAMS. I07 

a deflection 20% greater than for a corresponding uniform 
beam. 

Ill A. Uniform Strength and Uniform Breadth, — 

In these cases, as b is constant, I : I^ = /^^ : h^^ or I = I^ — 

V. Ream fixed at one end only, and loaded with W at 
the other. § 84, Fig. 28. 

((""v W _ ,.. ,. , , h 






^ ~ -^^ h\ 



dv W/^ 



dx E 



[_2 (/_^)^+ C] 



W/ ^ 1 1 

[ — 2 {i — xy + 2 /*] 



EIo 

3 



2W/ '2 [I 11 

Z' max. = ^ ^ — / r (/ — jc)* — / "^ 1 ^^ 
E lo J o '- ^ ^ 



or twice the deflection of a corresponding uniform beam. 

VI. Beam fixed at one end only and uniformly loaded 
with w per unit of length. § 84, Fig. 30. 

d'^v w (I — xY h I — X II — xY 

^ = - ^ fr^^ = -^ (log (/- -) -log /). 

V max. = • ^r^ / (log (/ — x^ — log /) dx * 
2EI0 J ^ 

= - \x log (/ X^ X — / log (/ X^ X log / 

wP wP 

(/ log o — / — / log o — / log / -[- /log /) = — 



-2EIo^ ^ ^^ ..^6^ I ^-^5"/ - 2E lo ^ 

or four times the deflection of a corresponding uniform beam. 

dx 
* Log (/ — x) zzzz ti \ dx ^ dv \ X z::i v\ dti '— — ' '' ' f log {J — '^) ^^^ 

i y y / y 

= X log (/ — x) 4- /— dx By division = — i -I- • .-. / (/x 

i// — X ' I — X I — X •/ / — X 

= — / dx -\- I j- = — X — / log (/ — Jt). 



Io8 STRUCTURAL MECHANICS. 

VII. Beam supported at both ends and carrying W at 
middle. § 84, Fig. 32. 

d'h> ^x ... ,„ ^ . h \2x ^ ^ ISx^ 



^^^ -" 2EI ^ °- 2^' • • //o - A/ / ~'°\7^' 



dv _ w/l r w/ & ^ 

TT — ■; — ~ — XT T \ X ^ dx ■-— . T^ T (2JC " -|- C). 

dx 4-1/2EI0J 41/2E lo ^ 



when ^ = A/, — - =: o, and C = — 2 \/i^- 

^ dx ^ ^ 



W/i /2 






7; max. = — — - / (x- — \/y)dx 

2y 2EI0 J o "^ 

^^^^ " ^ '17. F 



- 21/2EI0 ^^ Va^^j^ 

2-J/2E Iq V ^ 2-J/2 2-j/2y ~ 24E lo 



or twice the deflection of a corresponding uniform beam. 

VIII. Beam supported at both ends, and uniformly 
loaded with iv per unit of length. § 84, Fig. 34. 

-j^ =1 2E~I -^ ) ^ a : a Q^= X \l x) \ -^L ; 

_ 8|/^^"77~r^ ^ 
■ ° /^ 

dv wP r dx wP , . ~^ 2 X ^ ^. 
~T~ = ^T^ T / T = ^T^ T (versm -^^ \- C). 

^JC 16E I, / //. .2\i 16E In ^ / / 



{/x X^)^ 



dv -1 

— = o, when ^ = 1/ ; . •. C = — versin i = — \-, 



Zvl'^ n , . -1 2X 



V max. — : ^^ ^ / (versin — 4 77) ^jc 

i6EIo7 ^ ^ / 

Wp ~^ 2X ~1 

= ,5E J [ (^ —'40 versin — + i/(/^- — a"^) — ^ r: x ^ 



o 
7£//'^ r / TT "I 0.28^4 tt'/^ 0.0 t8 7£'/* 



[/ Ti "I 0.2854 7£'/^ 

2 4 J " 16E In ~ 



16EI0L2 4 J 16E Iq E Io 

or 37% greater deflection that for a corresponding uniform 

beam. 

Other beams might be analyzed, where both d and A varied at 
the same time. The method of analysis would agree with the 
above; but the cases are not of sufficient practical value to warrant 
their discussion here. 



DEFLECTION OF SIMPLE BEAMS. I 09 

112. Sandwich Beam. — If a beam is made up from two 
materials, placed side b}^ side, as when a plate of iron is bolted 
securely between two sticks of timber, the distribution of the 
load between the several pieces can be found from the con- 
sideration that they are compelled to deflect equally. As the 
spans and the longitudinal distribution of the loads are the 
same, the relationship W -=- E 1 i= W ^ E' V must exist, in 
which the symbols for one material are distinguished from 
those for the other b}' accents. If the depth is common, 
W ^ E ^ = W -=- E' d\ Since /"=: Mj, -^ I, the maximum 
unit stress will also vary as W -=- d. Therefore 

/=^or/:/' = E:E'. • 

If E for timber is 1,400,000 and for steel 28,000,000, the 
ratio of the stresses will be sV; and if /= 800 lbs. /"' = 16,000 
lbs. on the sq. inch. 

Example. — Two 4 in. X 10 in. sticks of timber, with a ^ X 
10 in. steel plate firmly bolted between them will have a value of 

^^ (800 . 8 -h 16,000 . 1) 10- • ^^ , 1 

M = ^ = 173,333 m. lbs., the plate sup- 
plying 1% of the amount. The combination, for a span of 10 ft., 

4M 

would safely carry = 5,778 lbs. load at centre, or 11,555 ^t)S. 

120 

distributed load, in place of 3,555 or 7,110 lbs. for the timber 

alone. 

113. Beams of Cement with Iron Rods. — Plates or 
beams of cement are used into which a sheet of wire netting 
or a combination of rods of iron has been built for the purpose 
of increasing the tensile resistance of the combination. They 
are known as Monier plates. The expansion and contraction 
of the two materials from changes of temperature are so 
nearly alike that heat and cold produce no ill effects. The 
ultimate strength of the combination lies at the yield point of 
the metal; for the expansion of the iron or steel above that 
limit is much greater than that of the encasing cement, so that 
the cement breaks. Rods of square section, twisted, are used, 
the twisted contour increasing the hold of the iron in the 
cement. 

The iron may be computed at not above 7, 500 lbs. per 
sq. inch, unit tension, neglecting the tensile resistance of the 



no STRUCTURAL MECHANICS. 

cement. The neutral axis, when the iron is placed at one- 
sixth the thickness of the plate from the tension side, should 
be assumed to lie at three-fourths the thickness from the same 
side. For mortar, one part of cement to three parts of sand, 
six months old, E = from 4,000,000 to 5,500,000, for the 
combination, a value higher than for pure cement. 

As to the comparative resisting power of brick laid in cement 
mortar, of concrete alone, and of concrete and steel combined, 
there may be cited the Austrian experiments of 1893 made on 
arches for bridges and floor surfaces. The stress at any point of 
any section of an arch is that due to a combination of thrust and 
bending moment, or the case of a strut-beam. The span was 
isVs ft- . 

Brick with cement mortar: — rise, 15^4^ in., thickness at crown, 
6 in.; breaking load 321.5 lbs. per sq. ft. 

Concrete: — rise, 15^ in., thickness at crown, 4 in.; breaking 
load 737 3 lbs. per sq. ft. 

Monier arch, concrete with wire netting 8 lbs. of steel per sq. 
ft.: — rise, 15^ in., thickness at crown, 3^ in.; breaking load, 
839.7 lbs. per sq. ft. 

Melan arch, concrete with I beams, 1.4 lbs. of steel per sq. 
ft.: — rise, 11 in., thickness at crown, 3^/^ in., breaking load, 3,360 
lbs. per sq. ft. 

The netting serves mainly to increase the beam strength in 
tension; the I beams, bent to the arch form and keyed against 
abutment beams, carry both the thrust and bending moment at any 
section, while the concrete assists the I beams by lateral support. 

114. Resilience of a Beam. — If a beam carries a single 
weight W, and the deflection under that w^eight is 7\, the 
external work done by that static load on the beam is ^Wv^. 
If this value of v is that which causes the maximum safe unit 
stress/", the quantity ^W7\ is known as the resilience of the 
beam, or the energy of the greatest shock which the beam can 
bear without injury, being the product of a weight into the 
height from which it must fall to produce the shock in ques- 
tion. For a beam supported at both ends, loaded in the mid- 
dle, and of rectangular section, v^ — —^-r^ ^^'^d W — — • 

6Eh 3/ 

fb/i^ fl- P 

Therefore, IWz;, = K I . -^-^ . f^, = tV . 4^ . bhi. 

I 6Eh E 

The allowable shock, or the resilience, is therefore pro- 
portioned io f^ -^ E, which is known as the modiilns of rest- 



i 



DEFLECTION OF SIMPLE BEAMS. Ill 

Hence of the material, and to the volume of the beam. These 
relationships hold for other sections, and beams loaded and 
supported differently. The above formula should not be rig- 
orously applied to a drop test, unless / is below the yield 
point. 

Example. — A 2in. X 2in. bar of steel, 5 ft. between supports, if 
y = t6,ooo lbs., ought not to be subjected, from a central weight, 

I 16,000^ . 2- . 60 o • 1U L 

to more than — - = ri8 m. lbs. 01 energy. 

18 29,000,000 

If the load is distributed over a similar beam, the deflec- 
tion at each point will be v, and the total work done will be 
\ f vivdx. If IV is uniform, and the beam is supported at its 
ends, 

Cl , w' fl rlx' x' Px~\ , w^P 

4 w / 7'ax = 



r/x^ X ^ ■^ \ 7 _ 

V 6 12 12 J 



4E I / ^ V, 6 12 12 7 240E I 

As = — , and v^ = :^ — , the above expression be- 

8 j/i Ej/, 

comes ^^{wl)v^, which is one-fifteenth of the value for a single 

load at the middle. 

115. Internal Work. — The internal work done in extend- 
ing and compressing the fibres of the beam must be equal to the 
external work yi^Nv^. 

Let the cross-section be constant. The unit stress at any 

point of a cross-section = /; the force on a layer zdy =■ pzdy. 

The elongation or shortening of a fibre unity of section and dx 

long by the unit stress/ = pdx -^- E. The work done in stretch- 

/- 
ing or shortening the volume zdydx = -J- . — - . zdydx. But / =: 

/ M 

— V = — - y. The work done on so much of the beam as is included 

between two cross-sections dx apart will be 



3- ~ — dx I y'^zdy = — ^^— f- ^^^• 

■;'i 



^ • EP— / " """ 2EI 



Substitute the value of M for a particular case, in terms of x, 
and integrate for the whole length of the beam. Thus for a beam 
supported at ends and loaded with W at the middle, M — ^ Wx 



112 STRUCTURAL MECHANICS. 

at any point distant .t from one end, for values of x between 
o and ^/. Then 

ax = / X ax 



2EI/ 4EI/ 96E I 

If this value is equal to the external work -|W27j, there results 

Vy = , as It should. 

40E i 

Example. — A weighted wheel of 1,000 lbs. drops ^ inch by- 
reason of a pebble in its path, at the middle of a beam, 3 in. x 12 
in,, 15 ft. span. If E = 1,400,000, to find / : 

External work = 1,000 . \ -\-\ . i,ooo6', = — -^ •^ . 12 . 180. 

"^^ '^ = T^eT^' 500(1 + ^^^^^f) ^ V^ • 

/ = 1,741 lbs. Resulting deflection = 0,56 in. Static 

unit stress would be 625 lbs., and z' = o. 2 in. In an actual 

bridge the shock is distributed more or less in the floor and 
adjacent beams. 

Examples. — i. What is the deflection at the middle of a 2 in. 
by 12 in. pine joist of 12 ft. = 144 in. span, supported at ends 
and uniformly loaded with 3,200 lbs.? E = 1,600,000. 0.27 in. 

2. What is the deflection if the load is at the middle? 

0.432 in. 

3. Find the stiffest rectangular cross-section, bh, to be obtained 
from a round log of diameter d. b ^ y^d. 

4. A 4 in. by 6 in. joist, laid flatwise on supports 10 ft. apart, 
is loaded with 1,000 lbs. at the middle. The deflection is found 
to be 0.7 in. What is E ? 1,607,000. 

5. What is the max. safe deflection of a 12 in. floor joist, 14 
ft. span, if/= 1,200 lbs. and IC 1,600,000. 

0.37 in. for uniform load; 0.29 in. for load at middle. 



CHAPTER VII. 

RESTRAINED AND CONTINUOUS BEAMS. 

ii6. Restrained Beams. — When a beam is kept from 
rotating at one or both points of support, by being built into 
a wall, or by the application of a moment of such a magni- 
tude that the tangent to the curve of the neutral plane at the 
point of support is forced to remain in its original direction 
(commonly horizontal) at such point, the beam is termed 
fixed at one or both supports. The magnitude of the 
moment at the point of support depends upon the span, the 
load and its position. It is the existence of this, at present 
unknown, moment which calls for the application of deflec- 
tion equations to the solution or such problems as those 
which follow, there being too many unknown quantities to per- 
mit the treatment of Chapter III. 

In applying the results obtained in the following cases to 
actual problems, one should feel sure that the beam is defin- 
itely fixed in direction at the given point. Otherwise the 
values of M, F and v will only be approximately true. 

117. Beam of Span 1, Carrying a Single Weight W 
in the middle and supported and fixed at both ends. Origin 

at left support. Fig. 52. 

The reactions and end moments are now unknown. The 
beam may be considered either as built in at its ends (as at 
the right in above figure), or as having an unknown couple or 
moment (^b applied at each point of support (as at the left), 
of a magnitude just sufficient to keep the tangent there 
horizontal. 

The reaction at either end will then be JW -f- Q, while 
the shear between the points of support will still be ± JW. 
For values of x < J/, 

M, = — Q(^ + ^) -h iW + Q) ^-^ = W^ — Q^- 



114 



STRUCTURAL MECHANICS. 



If this value is compared witli that of M^ in § 108, it is 
seen that a constant subtractive or negative moment is now 
felt over the whole span, in combination with the usual ^W^. 



- 4W/; 
Me = JW/; 

— IW/. 



M, 
M^ 
M, 



F. = iW; 

Fb = - iw. 




W/ 



-73 



V max. ^ 



IQ2E I' 



v^ — 



fh 



2 4Ey, 



S = wi »^^ - Q^^' £- = ri (^^^^ - Q^- + c)- 



— - =: o, when x =^ o; 
ax 



C = o. Also 



dv 
dx 



o, when r = \l; 



W/' 



o ^ 



Q^/ 



16 



; or - Q/. = - iW/, 



the negative bending moment at either end. That it is nega- 
tive appears by making :r = o in M^ above. 

If this value of Qd is substituted in the first equation, 
giving Mx = JW(^ — ^Z) the point of contrafiexure is located 
at Jt; = J/; and the bending moment at middle, where x = i/, 
is M max. = + ^W/, or one-half the amount in § 108. Sub- 
stitute the value of Qd in the equation for slope, 



//-) T W 

w 

V^ = — -^ (J x' — 1/x^ + C). 
As z^x = o> when x = 0; C == o. When x = -|/, 



z; max. 



~ ae IV24 16/ 1Q2E r 



The beam is therefore /our times as stiff as when only 
.supported at ends. 



RESTRAINED AND CONTINUOUS BEAMS. II5 

The slope is a maximum where — - or M^ = o, that is 

dx 

ian. max. slope = 



64E I' 



fl W/ ^^^ 8/1 , fP 

As-^ = — ; W = ^ and v, = ^^ ; 

so that only one-half the deflection is allowable that is 
permitted in § io8, but the beam may safely carry twice 
the load. 

It is useful to notice that this beam has a_ bending moment 
at the middle equal to that which would exist there, if the 
beam were cut at the points of contraflexure and simply sup- 
ported at those points; and that the two end segments, of 
length ^/, act like two cantilevers each carrying JW, the 
shear at the point of contraflexure. 

If the weight were not at the middle, the moments at the 
two ends would differ, equations would be needed for each of 
the two segments, and the solution, while possible, would be 
much more complicated. 

Example. — ^^A wooden beam, 6 in. square, and 7^ ft. span, 

is builtunto the wall at both ends. A central weight of 3,000 lbs. 

3,000 . 90 . 6 

will give a max. fibre stress of — -3 = 93714 lbs. per sq. 

o . 6 

in. at the middle and both ends. The deflection will be 

3,000 . 90^ .12 • -r T- -PL, 11 

— — — - = 0.07 m., it b = 1,500,000. ine ailowa- 

192 . 1,500,000 . 6 

1,200 . 90 . 90 . 

ble deflection, for/— 1,200, is = 0.09 m., and 

24 . 1,500,000 . 3 

max. allowable W = 3,000 . - — 3,860 lbs. 

7 

118. Beam of Span 1, Uniform Load of w per unit over 
the whole span, fixed at both ends. Origin at left support. 

Fig. 53- 

iwM ,, ^, iwl)l 
12 24 

(7C'l)P fP 

Fx = ^o(y — x). V max. = . v^ — — — -. 

^^ ' 384E I 32Ej'i 



ii6 



STRUCTURAL MECHANICS. 



As in the previous case, the reaction at either end may 
be represented by \ivl -f- Q. The shear at x is \zvl — wx^ 
which expression changes sign at the middle and at either point 
of support; hence at those places will be found M max. 

Mx = {^wl -f- Q) -^ — \wx^ — <^{b -\- x) = \wlx — \wx^ — Q<^. 

Compare with § 107. 



d'^v 



dv 
dx 



dv _ 
dx 

= o, when x 



— ^~ (^w/x^ — ^wx^ — Qi?x + C). 
hi 1 

dv 



C rz. O. 



C 



Al 



dx 



^ 



- Y(/''///^'7Z Z Zr/ \/^///////// 7777ZZ ZZZZZZZZZ 

i- -X- X -^ B 

I 



= 0, 


when X = I; 


luP 
4 


wP 

- 6 Q^^ = 


— 





M at middle 



V4 8 127 



the negative moment at each point of support. If j^ = 1/^ 

wl'^ 

.4 8 \zJ 24' 

Substitute the value of Q^, and get 

M, = —Iwix' — lx -\-\P)) 

\^x -^Lx -\- -^l xy, 



dv 
dx 



w 



w 



2E I 

A 



V = 



( X 

2E IVI2 



ix^ I ^x"^ 

+ -3^ + C). 



12 



Since v 



V max. 



o, when x ^^ o, C 

I ll 102 48 ^ 487 



o. When x = ^l, 






2E I\ 192 48 ' 48. 
which is one-fifth the value of § 107. 

The points of contraflexure occur where M^ = o; . •. 
x' — Ix -^ IP =: o; X = y ± hj Vl- 

The second term is the distance from the middle, each 
way to the points of contraflexure. If M is calculated for the 



RESTRAINED AND CONTINUOUS BEAMS. II7 

middle point of a span / -^- -1/3, it will prove to be zvl'^ -^ 24, 
as above. Since 

{wiy /I 12/1 fp 

^ — — , wl ~ — ; and v^ — 



12 j'l' y^l ' ' 32E;// 

or 0.3 as much as in § 107. The beam may safely carry, 
however, fifty per cent, more load. 

Example. — An 8 in. I beam, of 12 ft. span, carrying 1,000 lbs. 
per ft., if firmly fixed at both ends, and not to have a larger unit 
stress than 12,000 lbs., should have a value of I 

1,000 . 12 . 12 . 12 . 4 A o • -11 on 

= = 48. An 8 m. steel beam, 18 lbs. to 

12 . 12,000 

the foot, I = 57.8, will satisfy the requirement, -the load then being 

1,018 lbs. per ft. The deflection will be 0.6 in. 

< L -X I > 



'^^ ?////>^////////////Z 2 Z7Z///////////// 7 ^ //////^^^^ 



Pi ^- X ->( B p;^ try; ^ 



<- X -:^ D r/^.54. 




Pj = iwl. Pg = f^£'/. Mb = T%%WP. Md = — \wP. 

119. Beam of Span 1, fixed or horizontal at P2, sup- 
ported at Pi and carrying a uniform load of w per unit, Fig. 
54. Origin at P^ and reaction unknown. 

It will be seen from the sketch that a beam of length 2/, 
resting upon three equidistant supports in the same straight 
line, will come under this case. 

M, == Y,x — iwx'; ^ = ^j ii^i^' — i^^' + C). 

^ = o, when x = /;.'. C = i^^^ — h^/'- 
ax 

wPx Y^V^x 

-1- 
'4 

V — o, when jc = o; . • . C = o. \i x — I, v = o, and 
Pi = (,V — \)i<jl' -^ (i — i) /' = fe/. F, = %zol — wx. 
Substitute this value of P^ in the above equations. 



I rY^x^ wx^ wrx V^rx , ^,\ 



dxr 






J- = ^^'^' - '^' - '^^')^ ^' = sfi (^^^-^ - *-^' - *^'^)- 



Il8 STRUCTURAL MECHANICS. 

dv 
For V max., make — — =: o, or \lx^ — \x^ = i-^P. 
ax 

.-, x' — %lx^ = — i P. 

As a minimum value of v, or i; = o, occurs for x — l^ 
divide Zx^ — (^Ix"- -|- ^^ = o by .t — / = o, obtaining 

8jc^ — /jt" — /^ = o, or jc = / := 0.421^/. 

ID 

Then v max. = — 0.00^4 '^ . 

^ EI 

To find points of M max. put F^ = o, or jr = |/. 

Also, by inspection, M max. when x — /. 

For jc = f/, M max. = (ei- — ihYci'l'^ = ibze'/^. 

For X = I, M max. = (| — ^) zu/" = — |w/l 

For the point of contraflexure, ^/x — ij^^ = 0; or .t = |/; 
as was to be expected from the position of the point of max- 
imum positive M. 

Note again that the point of maximum bending moment 
is 710^ the point of maximum deflection. 

It will be seen that a continuous beam of two equal spans 
/, uniformly loaded with w per unit, has end reactions of ^za/, 
and a central reaction of 2 X %z^l ~- \zvl\ that points of con- 
traflexure divide each span at \l from the middle pier; and 
that the bending moment at the middle of the remaining seg- 
ment of f/ is, as above, I . f . \wP — ^^zvl'. It will also be 
seen that, since the bending moment at Pg is — \zvr\ a uni- 
form beam, continuous over two equal spans, each /, is no 
stronger than the same beam of span / with the same uniform 
load. It is, however, about two and a half times as stiff. 

Example. — A girder spanning two equal openings of 15 ft., 
and carrying a 16 in. brick wall 10 ft. high, of no lbs. per cubic 

r ^^^ ^ i i i- 5 • 4 • I ^ ^ • I O • 1 5 n_ i 

ft., will throw a load or = 2 7,i;oo lbs. on the 

.4-3 
middle post, and must resist a bending moment of 

4 . no . lo . 15 . 15 , ,, 

^ — ^ ^ = 41,250 ft. lbs. 

3 • ^ 
120. Two-Span Beam, -with Middle Support Low- 
ered. — A uniform beam, uniformly loaded, and supported at its 
ends, will have a certain deflection at the middle which can be cal- 
culated. If the middle point is then lifted by a jack, until returned 



& 



RESTRAINED AND CONTINUOUS BEAMS. 



119 



to the straight line through the two end supports, the pressure on 
the jack, by § 119, will be five-eighths of the load on the beam. 
Since deflection is proportional to the weight, other things being 
equal, — if the jack is then lowered one-fifth of the first deflection 
referred to, the pressure on the jack will be reduced one-fifth, or to 
one-half of the load on the beam. Hence, if a uniformly loaded 
beam of two equal, continuous spans has its middle support lower 
than those at its ends bv one-fifth of the above deflection, the mid- 
die reaction will be one-half the whole weight, the bending moment 
will be zero at the middle, and the beam may be cut at that point 
without disturbance of the forces. 

121. Beam of Span 1, fixed at left and supported at 
right end, and carrying a single weight W at a distance a 
from the fixed end, Fig, 55. Origin at fixed end. 



Mo — -p, a{l — a) {2I — a) 



M. 



W 



-(^ — ^) (3^ — ^)- 




Fo = -Tg a{l — a) {2I — a); 

The reaction at the supported end, being at present 
unknown, will be denoted by P.,, and moments will be taken 
on the right of any section x. From lack of symmetry, sep- 
arate expressions must be written for segments on either side 
of W. 



BETWEEN W AND FIXED END. 

Mx = Paa — a;) — W(a — X) 
dv 



dx 



= A[Po(te — Xx^) — Wlax — Vzxs) 4- C] 



r=A[P2r-^-^)-Wr^'-^) + C.r+C] 
\2 6^ V2 6/ 



2 6 
C = 0. 



dv , 

— — = o, when x = o; 
dx 

V ^= o when x = o; .". C — o. 

... C" = ^V{a' — ^a') 
If jc =: /, z; at Po = o: 



dv 
dx 



BETWEEN W AND SUPPORTED END. 

1 



= A[P2(lx-Ka;2) + C-] 



E I 



A. 



v^AlF.C:^-^) ^C X + C'] 
V 2 6 / 

dv , ^ dv 

U X = a, -^~ on left= -5- on right. 

dx dx 



It X — a, V on left = v on right. 



= hWa' 



U¥' — ¥') — iW^V + i\Ya' = o 



Wa- 



or p, = w^^ (1/ —1^) -- y' = -^ (3/ -- a) 



2P 



I20 STRUCTURAL MECHANICS. 

If this value of Pg is substituted in the above equations the 
desired expressions are obtained. Thus 



W«2 

Mx = -7r(3^ — a){l — x) — W{a — x) 






M max., by inspection, when ;ic ^ o, or :r = a. 

Wa'^ W 

M max. = — r^ (3/ — a) — Wa = ^ a{/ — a) (2/ — a), at 

fixed end; and = — ^ (/ — a) (3/ — a) at the weight. 

The point of contraflexure occurs between W and the 
fixed end, where M = o, or 

(3/ — a) (/ — x) — [a—x] = 0. . • . X =^ a/ 



2P^'' ^ ^ / V / 2/(/4- a) —a' 

di) 
The maximum deflection will be found where — = o, on 

dx 

on the right or left, according to the value of a. 

Various problems may be devised, such as those for finding 

values of a which will make -t-ti -1- or z^ a maximum, with the 

ax dx 

position of corresponding points of contraflexure and maximum 
deflection. They are more curious than useful. 

In solving the more intricate problems in the flexure of 
beams, as well as those just treated, each equation of condition 
can be used but once in the same problem, and as many unknown 
quantities can be determined as there are independent equations 
of condition. The reactions and moments at the points of sup- 
port are usually unknown, and must be found by the aid of such 
flexure equations as have just been used. 

The above beam may be regarded in the light of two 
equal continuous spans with W on each, distant a each side 
of the middle point of support. 

Example. — A bridge stringer which is continuous over two 

successive openings of 12 ft. each, and carries a weight from the 

wheels of a wagon, of 3,000 lbs.- at each side of and 3 ft. from 

the middle support, will be horizontal over that support. Then 

^,000 
— M max. — — -2 .3.9.21 = — 5,906.25 ft. lbs. -|- M 

S?000 „ r 11 T^ 3jOOO o 

max. = ^ 3 3' . 33 • 9 = 2,320.3 ft- lbs. P^ = ;^-— ^ . 3 • Zl 

2.12 2.12 

= 258 lbs. Reaction at middle support from both spans = 

2(3,000 — 258) = 5,484 lbs. 



CONTINUOUS BEAMS. 121 

CONTINUOUS BEAMS. 

122. Clapeyron's Formula, or the Three Moment 
Theorem for Continuous Loading. — To find the reactions, 
shears and bending moments for a horizontal, uniform, con- 
tinuous beam, loaded with zfj, lUo, w^, etc., loads per running 
unit over the successive spans /j, li, /g, etc. Fig. 56. P^ , 
Pi, P2, etc., denote the unknown reactions; Mq , Mi, M2, etc., 
the unknown bending moments at points of support, A, B, 
C, etc. ; Fq , Fi, F2, etc., denote the shears immediately to the 
right of A, B, C, etc.; while Y\, F'2, etc., denote the shears 
immediately to the left of the points of support B, C, etc. 

The origin of co-ordinates is first taken at A, and the 
supports are on a level. + M makes the beam concave on 
the upper side. As positive shear acts upward at the left of 
any cross-section, w is negative. 

Consider the condition of equilibrium of the first span 
A B, or /i, loaded throughout with zi\ per unit of length. 

?\>yyyyyyyyyl /^yy^y^/yy y^y////////////////y ////////////////^ N X 




^^ ^ > \P, f\Q.56. Vi ^nt 



Take moments on the left side of and about a section S, 
distant x from the origin A. The bending moment at S 
will be, by § 68, 

M = E I ^' z= Mo + Fo ^^ — \w,xK ( I.) 

ax" 

Let z'o , /j, i^ . . . i^ = tangent of inchnation of the 
neutral axis at A, B, C...N. Integrate (i.) between the 
limits o and x, transposing the slope i^ for the hmit zero to 
the left hand member, and thus obtain an expression for the 
difference in slope or inchnation of the two tangents to the 
bent beam at A and S. 



y^dx 




4 j 


= 


Mo 


X -j- 2^0^ 


— ^w,x' 


When X = 


= /i, 


dx 


— 


h> 


and hence 




EI(/i- 


- ^ 


) = 


M 


oh 


_L IF / 2 


iwj,\ 



(3-) 



122 STRUCTURAL MECHANICS. 

Integrate (2.) and determine constant as zero, because 
V = o, when x = o. 

E I (z; — 4 x) = iMo X' + iFo x' — hw,x\ (4. > 

Make x — \\. then v = Vy = o, and 

— E I 4 ^1 = iMo ^;^ -f iFo I,' — ^V7£/i/iS or 

— E r 4 = iMo ^1 + iFo /,^ — ^w^^. (5. ) 

Eliminate /q by subtracting (5. ) from (3. ). 

E I /, = iMo k + JFo i;' - ^w,l,\ ( 6. ) 

If the origin is taken at B instead of A, an equation Hke 
(5.) is obtained for the second span 4, or 

- E I /, = i M, 4 + i F, Z/ — i^w, //. (7.) 

Add (6.) and (7.), obtaining 

o = i Mo /j + i M, /, + i Fo I,' + i Fj // — i w, I,' — jV w, q. (8. ) 

The unknown slopes have thus been eliminated. The 
next step is to remove either M or F. Equation (i.) must 
equal M^ for x — l^\ therefore 

M M 

M, = Mo + Fo l,~\ w, k\ or Fo = ^-^ ° -\-\w, I,. 



In the same way, for 2d. span, F^ = — ^-^ -^ H~ ^ ^2 ^2- 



M, — Ml 

Substitute the values in (8.) and obtain 
Mo Z, , yi,l. Ml — Mo w,l,^ M2— Ml w^li 

2 2.3 6 6 12 



Wj ^1^ Wg h 



8 24 

Mo /i + 2 Ml (^1 + y + M2 ^2 = — i (^1 ^1' H- ^2 4')- (9- ) 

which is Clapeyron's formula for pier moments for a contin- 
uous beam, with continuous load, uniform per span. Notice 
the symmetry of the expression. The negative sign to the 
second member indicates that the bending moments at points 
of support are usually negative. 

Exa7nple. — Three spans, 30 ft., 60 ft. and 30 ft. in succession. 
Load on first and last 500 lbs. per ft., on middle span 300 lbs. per ft. 
No moment at either outer end. Then Mo =0. Mi = Mg by 
symmetry. 2M1 . 90 = — \ (500 • 3°^ + 3°° . 60^^). M^ 
= — 108,750 ft. lbs. Fo -= — 3,625 + 7,500 == + 3,875 lbs. 
F'l = -f 3,875 — 30 . 500 = — 11,125 lbs. Fi = 300 . 30 = 9,000 



CONTINUOUS BEAMS. I 23. 

lbs. .-. Po = P, = 3,875 lbs; P, = P, = 11,125 4-9'Ooo = 20,125 
lbs. The bending moment and shear at any point can now be 
readily determined. 

If the two adjacent spans are equal and have the same 
load, Mo + 4 Mx 4- M2 = — J zu/\ (10. ) 

If there are 71 spans, 7i — i equations can be written be- 
tween 71 -\- I quantities Mq, Mi . . . M^ . But it the beam is 
simply placed on the points of support, the extremities being 
unrestrained, Mo = O and M^ = o, and there remain 7i — I 
equations to determine Mj . . . M^ i- If the beam is fixed at 
the ends, the equations i^ = o and i^ = o will complete the 
required number. 

123. Shears and Reactions. — As the shear is the first 
derivative of the bending moment, § 68, from (i.) is obtained 

-^ = F ^F, — w,x, (II.) 

as was to be expected, -|- F acting upwards on the left of the 
section. A similar equation can be written for each span. 

The reaction at any point of support will be equal to the 
shear on its right plus that on its left with the sign reversed. 
As the shear on its left is usually negative, the arithmetical 
sum of Fjj and F'^ commonly gives the reaction, 

A simple example may make the application plainer. 
Given two equal spans, on three supports. 

w^ = iu.2 = IV. Mo = o, M2 = o. (10.) gives M^ = — ^wl'^. 
Fo = — ^ wl -)- 1 ^<;^ r= I ivl; F\ = ^ ivl — wl =^ — ^ wl. 

Fi = -J- loZ -f- 4 ^^^ ~ f '^^^ -'^'s ^ ^ wl lul =: § ivl. 

Pq = -I ivl; Pj = (t -[- I) wl = f ivl; P2 = I lul. 

I wl^ 

(5.) gives /o = — ^^ (o + tV — sV) ivl' = — ^^^^ . 

(6.) gives /j = — — (o -f A — i) 2(;P =: o 
and the analogous equation for the second span is 

which differs from 2"o only in direction of slope. 

(2.) gives E I ^ = \- t\ lolx- — . 

ax 48 6 

/ \ • T7 T ^^^^ I <? 7 3 ^-^^ 
(4.) gives K I z; =: — x 4- ^V wlx — — . 

48 24 



124 STRUCTURAL MECHANICS. 

These equations determine the slope and deflection at each 

dv 
point. Putting-— = o, there results V — 9 Ix"^ -\- '^ x^ ^ o, con- 
cix 

taining the root x ^ I, already known. Therefore divide by 

I — X = o, and obtain V^ -\- Ix — Z x^^ ■=- o, which is satisfied for 

x — 0.4215 Z, the point of max. deflection. The substitution of 

this value in the equation for v will yield v max. 

From (i.) M = | wlx — \ ivx'. 

If M = o, 1^ Z — ^ X ^ o, or X =i ^l, the point of contraflexure. 

Differentiate M, and get F = | loZ — wx. 

If F = o, jc = I /, the point of -)- M max. 

M max. =^ % wl . ^ I — - i ly . e V Z = tIs ^oP- 

Example. — If a uniformly loaded continuous beam covers 
iive equal spans, 

Mo + 4 M, + M2 = — i xd, = M, + 4 M3 + M3 
= M2 + 4 M3 -f M, =z M3 -f 4 M, + M,. 

Mo = o; M5 = o. Then M, = — i^ ivP = M^; 
M2 = — 3% wl^ = M3. 

Fo --3 If wl; F/ = —li ivl; F, = H wl; F/ = — H ivl; 

F2 = Hwl, etc. 

Po = Hwl = V,; Pj = Hwl = P,; P^ = U id = P3. 

The sum of the reactions must equal 5 tvl. 

124. CoefBcients for Moments and Shears. — It has 
l^een found that the numerical coefficients for moments and 
shears at the points of support, when all spans are equal and 
the load is uniform throughout, may be tabulated easily for 
reference and use. Thus the values of M and F just obtained 
for the five equal spans can be selected from the lines marked 
V. The reactions are given by the arithmetical addition of 
the shears. The sum of the reactions must equal the total 
load. The shears at the two ends of any span differ by the 
whole load on the span, the shear at the right end being 
negative. The dashes represent the spans. 

SHEAR AND REACTION COEFFICIENTS. 

\. \ — \wl 
II. -l — fl— |7^/ 

TTT 4 65 5 6 4 

^^^- 117 117 T 17 Tt7 TT7 1T7 

TV 11 1715 131 S 1517 11 

^^ • ^-^ 2-5" 2¥ 2^ ^-g- — 2F 28" — 2"8 

V IS 23 20 18 19 19 18 20 23 15 

etc., etc. 



CONTINUOUS BEAMS. I 25. 

PIER MOMENT COEFFICIENTS. 

II, — i — 7£^/2. 

TIT 1 1 w/2 

TV 3 2_ __ 3_ 

^^ • 2 8 ^8 58 

V 4 3 3 4 

^ • 5S g^¥ S"8 3^8 

etc., etc. 

The rule for writing either table is as follows: For an even 

number of spans, the numbers in any horizontal line are obtained 

by multiplying the fraction above, in any diagonal row, both 

numerator and denominator, by two, and adding the numerator and 

denominator of the preceding fraction. Thus, in the first table, 

2x6-^5 17 J. T i,T 2X1 + 1 -x 

r= — , and m the second table, ■ — — -^ , 

2 X 10 -f- 8 28 ' ' 2 X 10 4- 8 28 ' 

2 X "^ ~l~ 2 8 

or ^ — ■ = . For an odd number of spans, add the 

2 X 38 + 28 104 ^ ' 

two preceding fractions in the same diagonal row, numerator to 

1 "? —I— c; T 8 

numerator and denominator to denominator. Thus, — - — — 

28 + 10 ~ 38 

The denominators agree in both tables. A recollection of two or 

three quantities will enable one to write all the others. 

Example. — Continuous beam of 5 equal spans, each Z, carry- 
ing it» per ft. Where and what is the max. -[- M in second span. 
Shear changes sign at If I from left end of span. If this span 

were independent, -|- M at that point would be i- wP . '- 

19 . 19 

360 n^P ^' . . . . , 

= -— - . — — . Ihe negative or subtractive moment is ( jg- + ss . 
361-8 ^ ' 

ft) loP. The difference between these values is -)- M max. 

A more general investigation will produce equations which 
are of great practical value in the solution of problems con- 
cerning continuous bridges, swing bridges, etc., as follows: 

125. Three-Moment Theorem for a Single Weight. 

O is the origin, Fig. 57; the supports are at distances /i 
below the axis of x. A single weight W^ is distant /^4 from 
O on the span 4 , k being a fraction, less than unity, of the 
span in which W is situated. 

The moment at section S beyond W^ will be, as in the 
former discussion, 

M,= M„ + F„x — W„(^ — .f-4)- (r.) 

li X — 4 > Mx = M n+j , ^nd from ( i. ) 
M M 

Fn= ""'^ ^"^" +W„(I-^). ^,„y 



126 



STRUCTURAL MECHANICS. 



For an unloaded span, W = o, and Fj 



M 



m -M 



M, 



/. 



For the shear on the left of a section at the right end of the 
nth span, 



F' 



n + l 



w. 



M 



n + i 



M 



L 



For an unloaded span, W =r o, and F\ 



Mn, — M^. 

'in — 1 



As F'n, is the shear at left of support m, and F^ is the 
shear at right of the same support, the reaction there will be 



the sum of F^ and 



or 




ii^ 



dx = I Mx dx = Mn / dx -{- F^ 

/x 
(x — kL )dx. 
kin ^ ^ 



tA- Cfr^ lA' 



(2.) 



Note that the integral of the last term is between limits 
^4 3-rid X only. 

' dv 



E I 



(^ - 4 ] = M, ^ + 4Fn x' — iW„ (x - Z'4 r- (3.) 

Since the origin is at a distance /i^ above the ;/th support, 
the constant for the next integration is /i^ . 

El(v -inX — /in ) = 4M„ X' + IF, X' — JW, {x — kl^ )^ (4. ) 

which is the general equation of the curve of the neutral axis, 
the term in W disappearing for values of x less than /-/„ . 

\i X — l^, V — /^n + i-, If the value of F, from {\a) is 
inserted in (4. ), the slope at support ;/ is 



2n = 






(5.) 



CONTINUOUS BEAMS. 12 7 

The equation of the curve is therefore completely deter- 
mined when M^and M^ + i are known. The equation of this 
curve, between W^ and the n -j- ith. support, is ^iven by (4.), 
and the tangent of its angle with the axis of x by (3.)- I^ 
the value of F^ from (i^. ) and of i^ from (5.) are substituted 

in (3.), and x = l^, —- will be the tangent i ^^^ at ;/ + i» or 

ax 

Remove the origin from O to N, and derive an expression 
for 2nby diminishing the indices. 

Equate with (5.) and transpose. 

= 6E I ('Ipi^ + i^i±r^]_ w _^ /._^ ^k - ^^0 

which is the most general form of the Three Moment Theo- 
rem for a girder of constant cross-section. 

k —k^ z-^ k(i — k){i + k); 2k — ^k' -\- k' = k{l —k){2—k). 

Pier moments are usually negative and the end moments 
zero. When the supports are on a level h^ = Ji., etc. , and the 
term in E I disappears. 

Any reaction P^ = F^ — F'n; . *. 



n— 1 



See also Greene's Graphics, Part II., Bridge Trusses, 
Chap. ATII. 

Example. — Three span continuous girder as shown, carrying 
1,000 lbs. in first span and 2,000 lbs. in second span, at points 
indicated. Supports on a level. 



128 STRUCTURAL MECHANICS. 





O O 
O O 

, o , o 

O 30 „- I 20 ^- 2 3 




A 50' A 100' A 50' A 




300 Mj + 100 M2 = -1,000 . 2,500 . ID . TuV 




2,000 . 10,000 . T^iT . T% . rf 




100 M] -|~ 300 ^2 = — 2,000 . 10,000 . 1% . i%%. 




i(M, + M2) = — 13,200, -JCMi — M2) = — 7,200, 




Mj = — 20,400 ft. lbs. M2 = — 6,000 ft. lbs. 


Po 


_ 20,400 1,000 . 4 _ s lb- P — ^'°°° "^ 20,400 

50 10 ""' '■ LOO 




20,400 , 2,000 . 8 , 1,000 . 6 

+ .^ + T^ + T^ ^ ^'752 lbs. 
50 10 10 




^ 6,000 , — 20,400 -\- 6,000 2,000 .2 ^ ^, 

Pn = h h = 376 lbs. 

^ 50 ' 100 ' 10 ^' 




^ 6,000 

Po - — 120 lbs. 



50 

Examples. — i. A brick wall 16 in. thick, 12ft. high, and 32 ft 
long, weighing 108 lbs. per cubic ft,, is carried on a beam sup- 
ported by four columns, one at each end, and one 8 ft. from each 
end. Find M at the two middle columns, the reactions, and the 
value of I, if/ = ib,ooo lbs. 

M = — 31, 104 ft. lbs.; Pj = 3,024 lbs.; P^ = 24,624 lbs. 

2. Two successive openings of 8 ft. each, are to be spanned. 
Which will be stronger for a uniform load, two 8 ft. joists end to 
end, or one 16 ft. long? Find their relative stiffness. 

3. A beam of three equal spans carries a single weight. What 
will be the reactions and their signs at the third and fourth points 
of support, when W is in the middle of the first span ? 

— .%W; + ^W. 

4. A beam loaded with 50 lbs. per foot rests on two supports 
15 ft. apart and projects 5 ft. beyond at one end. What additional 
weight must be applied to that end to make the beam horizontal at 
the nearer point of support? 

156^4^ lbs. at the end, or 312^ lbs. distributed. 



CHAPTER VIII. 

PIECES UNDER TENSION. 

126. Central Pull. — If the resultant tension P acts along 
the axis of the piece, the stress may be considered as uniformly 
distributed on the cross-section S. If, then, f is the maxi- 
mum safe working stress per square inch for the kind of load 
which causes P, Fig. 58, 

P=/S;orS = P-^/, 

for the necessary section, which need not be exceeded through- 
out that portion of the piece where the above conditions 
apply. Changes due to connections will require a larger 
section. 

If, owing to lack of uniformity in the material, or the 
direct application of P, at the end of a wide bar, to a limited 
portion only of the width, the stress may not be considered as 
uniformly distributed at a particular cross-section, injurious 
stress may be prevented by taking the mean stress f at a 
smaller value, and obtaining a larger cross-section. 

If there is lack of homogeneity, or two materials are 
used together, or two or more bars work side by side, those 
fibres which offer the greatest resistance to stretching will be 
subject to the greatest stress. Fortunately, the slight yield- 
ing and bending of connecting parts tend to restore equality 
of action. 

A long tension member has a much greater resisting 
power against a suddenly applied load than a short one of 
equal strength per square inch. 

127. Eccentric Pull. — If the variation of stress on a 
cross-section is due to the fact that the line of action of the 
applied force does not traverse the centre of figure of the 
cross-section S, the force P' that can be imposed without 
causing a unit stress greater than / at any point in the section 
is less than P of the preceding formula, and depends upon the 



I30 



STRUCTURAL MECHANICS. 



perpendicular distance j/q of the action line of P' from the 
centre of S, 

For safe stresses, which must lie well within the elastic 
limit, the unit stress is proportional to the stretch, and plane 
cross-sections of the bar before the force is applied are as- 
sumed to remain plane after the bar is stretched. It is impos- 
sible to detect experimentally that this assumption is not true. 
Were the plane sections to become even slightly warped, the 
cumulative warpings of successive sections in a long bar ought 
to become apparent to the eye. No reference is intended 
here to local distortion preceding failure. 

If the stress on any section is not uniform and the suc- 
cessive sections remain plane, they must be a little inclined to 
one another. The stress on any cross-section S must there- 
fore vary uniformly in the direction of the deviation of the 
action line of P' from the centre, Fig. 58, and be constant on 

lines at right angles to 
that deviation. The 
stress on each particle 
may be divided into 
two parts, the mean 
stress, which is always 
the existing stress at 
the centre of the sec- 
tion, and the variable 
force P'; but the 
, must be balanced 



P^ Tig 58. 

part. The mean stress balances the 

moment of P' about the centre, or Pj/ 

by the moment of the variable part of the stress, taken 

about the axis in the plane of and through the centre of the 

cross-section, perpendicular to jo • Take the origin at the 

centre. 

Let / = unit stress at the point distant y from the centre, 
measured in the direction y^ and y^, which latter is the distance 
to the edge where the unit stress is f. If p^ = mean unit 
stress found at the centre, / — p^ will represent the variable 
part found at the distance y. 




P — Po -f—Po = y -yi, or/ —po 






y- 



PIECES UNDER TENSION. 131 

If ^ = variable width of section at distance y from the 
centre, the moment of the variable portion of the stress about 
the axis Z Z through the centre will be 

M = / y . zdy . y = ■ / y zay = I„ , 

where I^ denotes the moment of inertia of the cross-sectional 
area about the axis Z. 

This expression is the resisting moment of the cross- 
section. But M = P>o , and /^ = P' -^ S. . • . 

P'~\Iz. _ p^_ /S '_ /S 



^>°=(^-9f^^ 



r P 



I r'^ 

where r^ = I^ -^ S, to be measured in the direction j/j. 

Also/" = — -| I -|- '^^^ ), which gives the max. unit stress 

due to P' and y^ . 

Example. — A square bar, i in, in section, carries 6,000 lbs. 
tension. The centre of the eye at the end is \ in. out of line. 
Then/ =: 6,000(1 -\- \ • \ • 12) = 15,000 lbs. per sq. in., 2J times 
the mean and probably the intended stress. 

A bar which is not perfectly straight before tension is 
apphed to it, tends to straighten itself under a pull, but the 
stress will not become uniform on a cross-section. The bar 
is weaker in the ratio of p^ to f, as it might carry /"S if the 
force were central, but now can safely carry only p^ S. If a 
thrust is applied to a bent bar, there is a tendency to increased 
deviation from a straight line, and to an increase in the varia- 
tion of stress. 

It is seen from the example above, that a small deviation 
y^ will have a decided effect in increasing /"for a given P', or 
in diminishing the allowable load for a given unit stress. 
Herein may be the explanation of some considerable varia- 
tions of the strength of apparently similar pieces under test; 
and, on account of such effect, added to other reasons, allow- 
able working stresses may well be and are reduced below 
what otherwise might be used. 



132 STRUCTURAL MECHANICS. 

128. Hooks. — The bending action on and the strength of a 
hook are given by the same formulas. Here y^ will be the dis- 
tance from the inside edge to the centre of the cross-section, and 
yo the distance from the action line of the load to the same centre. 
Then the max. unit tension 



f'i('+m 



Example. — A hook, the section of which in the bend is ellip- 
tical, I in. X f in., carries the link of a chain at a distance 
of \ in. horizontally, from the inside of the bend. Then S — 

y . ^ = i sq. in.; r' =: ± = -^, § 94, Y.; yo = i in. Then 

/= 2P(i -f I . i . 16) = 18P. If / =^ 8,000, P = 450 lbs.; 
if y" = 12,000, P = 650 lbs. Compare with the given section. 
The ordinate of the bend should be reduced as much as possible. 

129. Combined Tie and Beam. — If to a tension mem- 
ber transverse forces are applied, or if it is horizontal and its 
v^eight is of importance, the unit tensile stress on the convex 
edge, due to the maximum bending moment, must be added 
to the unit stress at that point due to the direct pull. The 
former, 

/' = ^ = — ^~, and the latter/" = -. 

y I ' . S 

But / = f -\- f" must not exceed the safe unit tension, 
and the needed section is, since I = Sr^ 



\_ r(Mmax^i \ 



f 

In this case the sections may vary, since the external 
bending moment M varies from point to point. 

If the piece is rectangular in section, as with timber, the 
formula may be written, 

r 6M , P ■ I /'6M , ^^ 

In practical calculation of such a rectangular section, if 
h is assumed, it is sufficient to compute the breadth to carry 
M and add enough breadth to carry P, when the combined 
section will have exactly / at the edge. 



PIECES UNDER TENSION. I33 

Example. — A rectangular wooden beam of 12 ft. span carries 

a single weight of 3,000 lbs. at the quarter span, and, as part of a 

truss, resists a pull of 20,000 lbs. \i f z=. 1,000 lbs., what should 

3, coo .3.9. 12 

be the section under the \vei2;ht? M max. ^ 

^ 12 

81,000 .6 7 7.i r.^ T/< 7 

= 81,000 m. lbs, = b/r = 486. It /z = 12, /? = 3.37. 

1,000 

20,000 
Also = 1.67. Entire breadth = 3.37 x 1.67 = 5.04. 



Section = 5 x 12 in. The same result is obtained by the formula 

I /'6 , 81,000 ~\ 

b = 1 ■ -)- 20,000 1. 

12 . i,oooV 12 7 

130. Action Line of P Moved towards the Concave 
Side. — It will be economical, if it can be donej in a member hav- 
ing such compound action, to move the line on which P acts 
towards the concave side. If there are bending moments of oppo- 
site signs at different points of the length, or at the same point at 
different times, such adjustment cannot be made. Ifj^ois made 
equal to ^ (M max.) -^ P, one half of the bending moment will 
be annulled at the point where M max. exists, and at the point of 
no bending moment from transverse forces an equal amount of 
bending moment will be introduced. The unit stresses on the 
extreme fibres at the two sections will be the same, but reversed 
one for the other. 

Example. — A horizontal bar, 6 in. by i in. section, and 15 ft. 
long, has a tension of 33,750 lbs. It carries 100 lbs. per ft. uni- 

rormly distributed. M max. = = 33, 750 m. lbs. 



-7 T 7 c o . 6 
.' . Jq may be made \ in. Then / from M max. =: — = 

4- s,62S lbs. on either edge. But/ from P = ^^'J^^ (i ± ^ ' \l \ 

= 5,625 (i ± 1) = 5,625 ±_ 2,812.5. Stress at top at ends and 
at bottom at middle — 8,437^ lbs.; at bottom at ends and at top 
at middle — 2,8i2|- lbs. 

The extreme fibre stress from bending moment of the load var- 
ies as the ordinates to a parabola, that from Yy^ is constant. A 
rectangle super-imposed on a parabolic segment will show the 
resultant fibre stress at each section. 

131. Connecting Rod. — If a bar oscillates laterally rapidly, 
as does a connecting rod on an engine, or a parallel rod on loco- 
motive drivers, there will be a force developed due to the accelera- 
tion, which force will tend to bend the bar as does a distributed 
load. Each particle of the bar will exert a lateral force (like cen- 

triiugal force ) or — . — or — ocv, where c< = 2~)i\ ?i = the num- 



<b 



r 



134 STRUCTURAL MECHANICS. 

ber of double oscillations of the bar, or number of revolutions, per 
second, of the crank; r = radius of crank, or amplitude of oscil- 
lation at the particular point. Then ii w =i weight of a foot in 
length of the bar, and ^ = 32.2 ft., the bar will suffer a bending 

w 
moment and a resulting fibre stress due to a load — att^^zV per unit 

cr 

of length, which stress must be added to the tensile stress due to a 
pull or to the compressive stress due to a thrust. The radius or 
amplitude r is constant for the parallel rod but varies uniformly 
along the connecting rod; w may be constant or vary. An I 
shaped section is suitable for such cases. Owing to the rapid var- 
iations and alternations of stress, the maximum unit stress should be 
small. Mass is disadvantageous in such rods. 

132. Tension and. Torsion. — A tension bar may be sub- 
jected to torsion when it is adjusted by a nut at the end, or by 
a turnbuckle. The moment of torsion will give rise to a unit 
shear at the extreme fibre, for a round rod, of ^ = T -^ ^'r^^ 
by § 89, or at the middle of the side for a square rod of ^ = T 
-^ o. 2o8>^^ by § 90, either of which, combined with/= P 
-7- S, the tensile stress, will give p^ = ^/ -\- \/{k/^ + ^^)- 

§93. 

Example. ^-Pv round bar, 2 in. diameter, to be adjusted to a 
pull of TO, 000 lbs. per sq. in., calls for the application to the turn- 
buckle of 200 lbs. with an arm of 30 in., one half of which moment 
may be supposed to affect either half of the rod. If the turn- 
buckle is near one end, the shorter piece will experience the 

3,000 .2.7 
greater part of the moment. ^ = 3 — = 1,9^0 lbs. The 

max. unit tension on the outside fibres of the rod will be 5,000 
+ 1/(5.000' -\- 1,910') — 10,350 lbs. 

133. Tension Connections. —If a tension member is 
spliced, or is connected at its ends to other members by 
rivets, the splice should be so made, or the rivets should be 
so distributed across the section as to secure a uniform distri- 
bution of stress. An angle iron used in tension should be 
connected by both flanges, Plate III. , if the whole section is 
considered to be efficient. One or more rivet holes must be 
deducted in calculating the effective section, depending on the 
spacing of the rivets. See § 217. If the stress is not uni- 
formly distributed on the cross-section, the required size will 

be found by §127, S = -Y i + "^-^ V 



PIECES UNDER TENSION. I35 

Transverse bolts and bolt holes are similar to rivets and 
rivet holes. 

Timbers may be spliced by clamps with indents, and by 
scarfed joints, Plate II. , in which cases the net section is 
much reduced; so that timber, while resisting tension well, is 
not economical for ties, on account of the great waste by 
cutting away. However, where the tie serves also as a beam, 
timber may be very suitable. 

134. Screw Threads and Nuts. — If a metal tie is 
secured by screw threads and nuts, the section at the bottom 
of the thread should be some fifteen per cent, larger than the 
given tension would require, to allow for the local weakening 
caused by cutting the threads. Bars are often upset or en- 
larged at the thread, to give the necessary net section, and 
thus save the material which would be needed for an increase 
of diameter throughout the length of the rod. 

To avoid stripping the thread, the cylindrical surface, 
whose area is the circumference at the bottom of the thread 
multiplied by the effective thickness of the nut, should be, 
when multiplied by the safe unit shear, at least equal to the 
net cross-section of the rod multiplied by the safe unit tension. 
2 T. r'.t.q = r, r'^j\ or/r' = 2 q t. As q is usually taken 
less than f\ as with a square thread only half of the thickness 
is effective, and with a standard V thread quite a portion of 
the thickness must be deducted, nuts are usually given a thick- 
ness nearly or quite equal to the net diameter of the rod. 
Heads of bolts may be materially thinner. 

135. Eyebars. — The eye for the connection of a tension 
bar to a pin is seen in Fig. 59. The pin is turned and the 
eye bored to a reasonably close fit. Since bearing first takes 
place at the back of the pin, the most intense pressure will be 
found there, and it will probably diminish at different points 
of the semi-circumference until the horizontal diameter is 
reached. The pressure on the pin may be found to extend 
slightly below that diameter. If these pressures are assumed 
to be normal, and are laid off in succession at 2'-2, 2-3, 3-4, 
. . . 5-6, and closed with 6-6', the pull on the bar, a pole can 
be assumed at o and an equilibrium polygon or curve drawn, 
cutting the dotted centre line of the material of the eye about 



136 



STRUCTURAL MECHANICS. 



as shown. By moving o horizontally and changing the point 
of beginning near A vertically, this equilibrium curve can be 
brought to the right position to satisfy the requirement that 
the sum of the products, from A to D, of each ordinate nor- 
mal to the equilibrium curve multiplied by the force 0-2, 0-3, 
etc., there acting, shall equal zero. This requirement means 
that the tangents at A and at D, to the centre line, shall make 
the same angle with each other, before and while the pull is 
applied. See Greene's Graphics, Part II., Chap. VI. 

It will be seen that the resultant force o-i at the section 
at A is smaller than 0-4 or 0-6, the pull at B or C. Hence 

. considerable deviation of the result- 
-f^^ ant force from the centre line at that 
section is not a serious matter. 
The eye was formeily made with an 
unnecessary enlargement at A, but 
is now commonly made circular 
through more than half of its peri- 
meter. The edges of greatest stress 
are at A on the outside, B on the 
inside and C on the outside. This 
neck should be wide. The material 
in front of the pin within the dotted 
triangular area is of no service. In the looped eye of Fig. 
59, made by bending the bar around the pin, that space is 
empty. Experiment has shown that for strength this loop 
should be long, from two to two and one-half diameters of the 
pin. If it were not for the weld and the excess of metal on 
either side of the pin, such a form of eye would be a satis- 
factory one. 

If Jo is the deviation of the force from the centre of sec- 
tion at A or B, Fig. 59, the half width being jj and the pull 
o-i or 0-4 being P, the unit stress at the extreme edge, by 
§ 127, will be 

/.£(. + .-). ,.(. + 'z-). 

where b = width of eye on one side. 

It is necessary to make the pin from |- to J the width of 
the bar, in order to develop the strength of the latter, that is 




PIECES UNDER TENSION. I37 

to give sufficient bearing or compression area back of the pin. 
The right section through the eye exceeds that of the bar from 
33 to 50 per cent. 

Examples. — t. A round bolt i^ in. in diameter carries a load 
of 20,000 lbs. As its head is not square to its length, the centre 
of resistance is probably \ in. from the axis of the bolt. What is 
y"in this case and how much greater than the mean stress? / = 
41,500 lbs. Since the elastic limit has been passed the actual 
maximum stress is probably less. 

2. Find the maximum stress and the mean stress for a pull of 
20,000 lbs. on a square eyebar \\ in. X ^\ in. if the pin is \ in. 
out of centre. 26,670 lbs. 

3. A rectangular bar, section 2 in. X i in., has a central pull 
of 8,000 lbs. Then/" = 4,000 lbs. If the bat is widened to 3 in. 
without change of force and its point of application, what is/"? 

4,667 lbs. 



CHAPTER IX. 

COMPRESSION PIECES. COLUMNS, POSTS AND STRUTS. 

136. Blocks in Compression. — If the height of the 
piece is quite small as compared with either of its transverse 
dimensions, and the load upon it is centrally imposed, the 
load or force P may reasonably be considered as uniformly 
distributed over the cross-section S, and the unit stress /"upon 
each square inch of section will be given by the formula 

P=/S, or/= P- S, 

as is the case with any tension member when the force is 
centrally applied. 

137. Load Not Central. — So also, when the action line 
of the resultant load cannot be considered as central, but 
deviates from the axis of the piece a distance j/q , the force P 
can be replaced by the same force acting in the axis and a 
couple or moment Fj/^ , which moment must be resisted at 
every cross-section by a uniformly varying stress, forming a 
resisting moment exactly like that found at a section of a 
beam. Compare Fig. 58, and change tension to compression. 

If /' is the uniform unit stress to resist the central load, 
so that/' = P -^- S; and/" is the unit stress at the extreme 
fibre which lies in the direction in which j/q is measured, and 
at a distance j', from the axis, 

P)'o=/"I-J. orr = Fyoy,-^ I. 
Hence the unit stress on the extreme fibre on the com- 
pression side, /. e. , on the side towards which y^ is measured, 
will be, since I = Sr^, 

/ = /' + /" = I + ^ = |(i + ^-^) =/'(>+ ^-7#^j^ 

The load that such a piece will carry is 

/S 

P = —j: • 



COLUMNS. 139 

By comparison with the formula of the preceding section, 
it will be seen that the piece, when the load is eccentric, is 

weaker in the ratio i : i -j- •; \ 



r' 



The values of j/^S -^ I or j/j -^ r'^ are given below for some 
of the common sections of columns, ji being measured in the 
direction /i. 







I. 


yv 


s. 


y,S - I. 


Rectangle, 




12 


y2h 


bh 


6 


Square, 




12 


%h 


h' , 


6 

I 


Circle, 




64 


y2d 


ji^d' 


8 


T-T <^ 1 1 (^ Axr T?(:i/^fTn 


, bh^ 


— b'h'^ 


\/h 


hh h'h' 


6/i{bA b'h') 



^ ' 12 /" bJl' — b'h'^ 

Hollow Circle, '{d' — d") y i^ .^2_^m '^^ 

138. The Middle Third. — The mean or average unit 
stress is always found at the centre of the cross-section. 
When the maximum unit stress at the extreme fibre becomes 
twice the mean stress, the stress at the opposite edge, if the 
centre is in the middle of h, drops to zero. That is /" = /' or 
j^oj^i -^ r^ = I. This will occur when y^ = r^ ~ y^, y^ then 
being, for the rectangle, \Ji, and for the circle, \d. Hence, for 
a rectangular section in masonry, the centre of pressure must 
not deviate from the centre of figure more than one-sixth of 
the breadth in either direction, if the unit stress at the more 
remote edge is not to be allowed to become zero or tension. 
As masonry joints are supposed, in many cases, not to be 
subjected to tension in any part, the above statement is equiv- 
alent to saying that the centre of pressure or line of the 
resultant thrust must always lie within the middle third of 
any joint. 

Likewise, for two cylindrical blocks in end contact, the 
centre of pressure should fall within the middle fourtJi of the 
diameter, if the pressure is assumed to be uniformly varying 



140 STRUCTURAL MECHANICS. 

and it is not permissible to have the joint tend either to open 
or to carry tension at the farther edge. 

The unit pressure at the most pressed edge of a rectangle 
can be found for any deviation j/^ . 

1st. When the stress is over the whole joint, as before, 



/=IO+^-^> 



2d. When compression alone is possible, and only a part 
of the surface of the joint is under stress. The distance from 
the most pressed edge to the action line of P is \h — y^. 
The entire pressed area = 3(i/^ — Jo)^^ since the ordinates 
representing stresses make a wedge whose length along y 
is three times the distance of P from the most pressed edge. 

P = */• 3(i/^ -yo )b\ f= l-P -^ {Wi -y, )b. 

If the case is that of a wall, and P is the resultant force 
per unit of length, b =. \. 

As j/q increases, /increases, until finally the stone crushes 
at the edge of the joint, or shears on an oblique plane as 
described in § 23. Sometimes the pressure is not well dis- 
tributed, from poor bedding of the stones, and spalls or chips, 
under the action of the shearing above referred to, may break 
off along the edge, without failure being imminent, since when 
the high spots break off others come into bearing. 

P can never traverse one edge of the joint, if tension is 
not possible at the other edge, as the unit stress then becomes 
infinite. Some writers commit an error in determining the 
thickness of a wall by equating the moment of the overturning 
force about the front edge or toe with the moment of the 
weight of the wall about the same point. This process is equiv- 
alent to making the action line of P traverse that point. The 
centre of moments should be taken either at the outer edge of 
the middle third of the joint, when pressure is desired over the 
whole joint, or about a point at such a distance \Ji — y^ from 
the front as will give maximum safe pressure at the front edge. 
A uniformly varying stress extending over three times that dis- 
tance will equal P, as lately stated. A portion of the joint at 
the rear will then tend to open. 



COLUMNS. 141 

Examples. — A short, hollow, cylindrical column, 12 in. exter- 
nal diameter, 10 in. internal diameter, supports a beam which 
crosses the column 2 in. from its centre. 

8^ _ 8 . 12 96 _ P r 2 . 96 



b V. 244 J 



d^ -f~ d 144 + 100 244 S V 244 

= /o (i ± 0-8), or the stress at either edge will be 80^ greater 
and less than the mean stress. 

A joint, 10 ft. broad, of a retaining wall is cut at a point 3 ft. 
9 in. from the front edge by the line of the resultant thrust above 
that joint. If this thrust per ft. of length of the wall is 28,000 lbs., 

1 r- 1 r 1 -n 1 28,000. 

the pressure per sq. rt. at the rront edge will be (i -j- \\.-ii) 

= ij . 2,800 = 4,900 lbs. At the rear it will be 700 lbs. per 
sq. ft. 

139. Resistance of Columns. — A column, strut or other 
piece, subjected to longitudinal pressure, is shortened by the 
compression. As perfect homogeneousness does not exist in 
any material, the longitudinal elements will yield in different 
amounts, so that there is apt to be a slight, an imperceptible 
tendency to curvature of the strut. Hence the action line of 
the applied load may not traverse the centres of all cross-sec- 
tions of the piece. The product of the applied force into the 
perpendicular distance of its line of action from the centre of 
any cross-section will be a bending moment, which must 
develop a resisting moment at the cross-section, resulting in a 
varying stress, as in § 137. Equilibrium of the column as a 
whole will occur only when, for a given load, the axis of the 
column has assumed such a curve that, at each cj^oss-section, 
the resisting moment against lateral flexure equals the bend- 
ing mo7nent at that point due to the external force . If the 
load is too great for that condition to be fulfilled, failure by 
flexure takes place. 

140. Remarks. — This curvature under longitudinal pres- 
sure can be readily obtained with test specimens of most 
materials, even with some samples of cast-iron, and the form 
of the curve apparently conforms to the one to be deduced by 
theory. A tendency to such a curve must therefore exist 
under working stresses, although the curvature is impercep- 
tible, unless the column happens to have its load perfectly 
axial, a contingency that cannot be safely relied upon. The 



142 STRUCTURAL MECHANICS. 

column formula, so-called, should therefore be confidently 
applied. 

Further, as such curvature can be produced in test speci- 
mens not more than four or five diameters long, (Plate I, 
right figure, cast-iron; left figure steel), such a formula is ap^ 
plicable to columns and struts of any length. It is not nec- 
essarily to be applied, however, to very short posts, or blocks, 
for the relation P = /S will determine their size with sufficient 
exactness. 

141. The Yield Point Marks the Column Strength. — 
The influence of j/^ in determining the load a compression 
piece will carry has been shown in § 137 to be very marked. 
A column which has become sensibly bent under a load is very 
near complete failure. The moment of the load at the cross- 
section of greatest lateral deflection has then become so large 
that the stress on the extreme fibres passes the yield point, 
and the great increase of stretch at and above the yield point 
at once increases the bending moment greatly. Hence it is 
true that the yield point marks nearly the ultimate compress- 
ive strength of materials when tested in column form. § 149. 

Again, the fact that, in tests of large columns, a zje?y 
slight shifting of the point of application of the load at 
either end has a decided influence on the amount of weight 
such a column will carry is a confirmation of the statements 
with which this discussion opened. It also has a bearing upon 
the truth of the theoretical deduction as to the effect of eccen- 
tric loading, as discussed in § 137, and to be applied to long 
columns later. 

142. Direction of Flexure. — The flexure usually occurs, 
unless there is some defect or weakness, in a direction /<^r<7//^/ 
to the least trmisverse dimension of the strut, i. e., perpen- 
dicular to that axis in the cross-section which offers the least 
resisting moment. By the application of longitudinal pressure 
to a slender rod its flexure may be made very apparent. The 
fomn of the column formula ought to resemble that of § 137, 

P =/S -- I +-^. 

143. Formula for Columns. The Flexure Curve. — 
If the column is fixed in direction at its ends, by its connec- 



COLUMNS. 143 

lions to other pieces, or by having a broad, well-bedded base 
and cap, it will act in flexure much as a beam fixed at the 
ends. A couple or bending moment, which may be repre- 
sented by Mq , will thus be introduced at each end. Let 
P = applied external force or load; v = any deflection ordin- 
ate, measured at right angles to the action line of P, from the 
original axis of the column to any point in the axis when bent; 
:x: = distance from one end along the original axis to any 
ordinate v; I = length of column. 

The combination of the moment Mq at the end of the 
column with the force P has the effect, as shown in Mechanics, 
of shifting the force P laterally a distance M^ -i-F = v^ ; hence 
the action line of P is now parallel to the original axis, at a 
distance v^ from it, or in the line F E of Fig. 60. The ordin- 
ate to the points of contraflexure is therefore v^ . This action 
can be more fully realized by conceiving that the bearing sur- 
face at A is removed, and that P acts at such a point on a 
horizontal lever as to keep the tangent to the curve at A 
strictly vertical. 

As V is measured from the original axis, the bending 
moment at any section is M = P (v^ — v), which will change 
sign when the second term is larger than the first. 

If the flexure is very slight, an equation similar to that 
used with beams may be written. 

Multiply by dv, and integrate. As — = 0, when z/ = o, 

dx 

When — = o, at U, the middle of the length, v max. = 2 v^, 
dx 

or double the ordinate at the point of contraflexure F. Let 

P -- E I = B. 

The square root of the above equation gives 

dv 



^ = 1/ (B (2 z;^ z^ - v^) ), or dx = ^ Q )- 



( 2 e^o z^ — z>^)' 



144 



STRUCTURAL MECHANICS. 



Integrate. 



As z/ = o, when x 

-1 v 



X 



a)( 



versm 



O, 



+ (C' = o)). 



v \/ 



i\ I 



V — Vq versin {x -|/B) = Vq [i — cos {x -j/B) ]'. 
As I — cos 6=2 sin'"^ ^6, v = 2 z'^ sin^ (^ x -j/B). 

If, in this equation, x = i/, a value of v max. is obtained 
;,-. to be equated with the previous value 2 v^. 

2 Vo — 2 v^ sin^ (1/-/B), or i = sin' (l/-,/B). 
sin (J/ |/B) = I, or J/ -j/B = ^ tt, since sin"^ i = ^ tt. 

47:' E I 



As B = P 



E L 



I' 



;^<:-ul 



which is commonly known as Euler's Formula. There 
fore the equation of the curve of the column is 



D 



2 Vr. sm'^ 



G-) 



To find the points of contraflexure, make v = v^ 
sin ( -^ ^ j = ]/|- = si] 



sin45 



sm 



\h 



A A 



X 

T 



=. ^ ox X 



\l from either end. 



i' Hence the curve is made of four equal portions,. 

Fi^M. A F, F D, D E and E H. 

A column hinged, pin-ended, or free to turn at its ends, 
and of length represented b}^ E F == J/, will have the same 
portion of stress at D that is due to bending as does a column 
of length A H = /, which is fixed at its ends. 

If, in actual cases, F is considered to be practically in the 
same position horizontally as before loading, it may be said 
that a column fixed at one end and hinged at the other, of 
length F H = |/, will also have the same portion of stress 
arising from bending. The maximum deflection will then 
occur at one-third of its length from the hinged end. This 
result has been verified by direct experiment on a full-sized 
steel bridge member. 

144. Formula for Load. — The load which a fixed column 
or strut will safely carry is determined by the maximum unit 



COLUMNS. 145 

stress on the extreme or outside particles on the concave side 
of the column at D, or at A and H, all three of which points 
are those of maximum unit stress. The unit stress here is 
limited to /, the maximum safe unit stress of the material, 
and the allowable load on a column is limited accordingly. 
This unit stress /in the extreme fibres maybe separated 
into two parts, one of which /' is the uniform compressive 
stress from P, or/' = P -^ S, and the other is/ — /' the 
maximum compressive stress in the extreme fibres on the con- 
cave side from the bending moment P^^^ . Were there no 
direct compression, the post or strut could safely carry P as 
given by P = 4-^E I -^ /-; but as/' exists at the same time, 
the load P must be reduced in the ratio of the available com- 
pressive stress remaining to resist bending, or/ — /', to the 
max. safe stress/ Therefore multiply the value of P by this 

ratio (/ — /) -/:=. '""f • 



P = 



/s /s 



, /ys ~ /■'' 

4- tj L r 

where r^ — \ -^ S, or what is known as the square of the 

■f 
radius of gyration, and ^ = , a quantity dependent upon 

the material. In practice, a is often greatly increased. See 
§§ 174-5. The last formula is known as Rankine's. 

This value of a agrees very well with the one given by Rankine 
for iron, although the constants in the latter case were derived from 
experiments carried to failure. As such failure occurs at but little 
above the yield point, the difference in results will not be large. 
\i f ■=! 36,000, 4-- = 40 approximately, E = 28 to 30,000,000, 
<^ = I -^ 31,000 to 33,000. 

Rankine gives 36,000 S -^ I i + ~. 5 I- 

V 36,000/-^ 

145. Multipliers of a. — As seen above, 2/ should theo- 
retically be substituted for / when the column is hinged or free 



146 STRUCTURAL MECHANICS. 

to turn at its ends, in order to obtain the equivalent length of 
a column which is fixed at the ends; and for a column fixed at 
one end and hinged at the other il should be substituted for /, 
for the same reason. Or, more conveniently, «'may be used 
for a column fixed at the ends and of length /; 4^ for a column 
hinged at both ends and of length /; and ^a for a column 
hinged at one end and fixed at the other, length /. Actual 
tests, carried however to the extreme of bending or crippling, 
appear to show that a column bearing on a pin at each end is 
not hinged or perfectly free to turn; hence the multipliers of a 
more commonly used, instead of i, ^ and 4, are i, I and 2. 
The theoretical ratio of strength of a column hinged at 
£nds to that of one fixed at ends is 



(^ + ^7.)- + 4^-0 = '-I 



-\- \aV 



\\ I — loor, the ratio becomes ; and, 11 a 

I + 40,000^ 

^ . 1 46,000 ^ 01^^ A 

it becomes ^ , or 23 to 38, about 0.6. A 



" 36,000 76,000 

column fixed at ends is, for the above values, some sixty-five 

per cent., or two-thirds stronger than one hinged at ends. 

146. Pin Friction. — Some regard columns as neither per- 
fectly fixed nor perfectly hinged, and use but one value of a for all, 
which might perhaps then be taken as a mean value. The moment 
of friction on a pin is considerable. If P is the load on a post or 
strut, d the diameter of the pin, and tan ip the coefficient of friction 
of the post on the pin, the moment of friction at the pin will be 
V . \d . tan ip; and this moment, if greater than Mq == Pz^o ? will 
keep the post restrained at the end, so that the tangent there to the 
curve remains in its original direction. As P and Vq increase, Mq 
will become the greater when v^ exceeds \d tan o\ the column will 
then be imperfectly restrained at its ends, and the inclination will 
change. As the friction of motion is less than that of rest, such 
movement when started, may be rapid. Some tested columns, 
showing at first the curve of Fig. 60, known as triple flexure, have 
suddenly sprung into a single curve and at once offered less resist- 
ance. 

It may be doubted whether ordinary columns, under working 
loads, ever develop a value of Vq sufficient to overcome the pin 
friction, unless the column is very slender and the diameter of the 
pin small. 



COLUMNS. 147 

147. Failure by Tension. — In rare cases, when the mater- 
ial has but moderate tensile strength, as compared with the com- 
pressive strength, for example cast-iron, and the column is very 
slender, the convex side may be the weaker. The maximum 
allowable unit tension arising from the bending moment will be 
— {/-{-/'), reducing to an actual unit tension — /"when the uni- 
form compression/"' is combined with it. The formula for P then 
becomes 

a — , — I 

r 

148. Short Columns and Slender Columns. — If a col- 
umn is very short, the second term in the denominator of the 
value of P, § 144, becomes very small, and the formula prac- 
tically reduces to P = /"S. Some engineers use this form for 
iron and steel built struts up to a limit of / = 6or or 8or. 

On the contrary, if the column is extremely slender, the 
second term of the denominator becomes so much larger than 
unity as to practically overpower the first term, and the 
expression becomes Euler's Formula, § 143. In that case the 
stress on the extreme fibre from the moment of flexure far 
outweighs that from the direct thrust, and the latter may be 
neglected. 

' Example. — A wrought iron column of hollow cylindrical 
form, 20 ft. long, not fixed at ends, must carry a static load of 

4 . 10,000 I 

^6,000 lbs. If / = 10,000 lbs. : 4^ = — 7y -=, ■ = —z ; 

^ -^ 4.7z\ 28,000,000 28,000 

and the mean diameter of a thin ring is d; then by VI., § 99, 

2 1 v9 A A 10,000 r^/ i6.5-\ 22 

^ = ¥', and 36,000 ,^^ , ,^, , 8 - ^^^-^(^ + ^ =7 



dL 



I -t- 



28,000 d 



If/ — lin., — d^ — — ^2= CO. 4. Solve for ^ by trial. As d 
^ 14 10 ^^ ^ 

is between 6 and 6 J in., a cylinder of 6 in. interior diam. and J in. 

thick will be satisfactory. 

149. Experimental Results. — The crippling strengths of 
different columns have been obtained experimentally for vari- 
ous lengths and values of S and r. When such results are 
plotted with ordinates P -^ S = / and abscissas / -^ r, and 



148 



STRUCTURAL MECHANICS. 



curves are drawn to agree as nearly as may be with the mean 
locus of the more or less scattered points thus found, it is not 
surprising that the equations of such curves do not agree very 
closely with the one deduced in § 144. The general trend of 
the area covered by the points is, however, reasonably satis- 
factory. The same reason for variation applies here as in the 
deduction of / from tests of beams loaded to rupture, in 
which tests a value of / is obtained differing from the ultimate 
tensile and compressive strength of the material. The 
assumption of plane sections and uniformly varying stress will 
be true only below the elastic limit, and only within that limit 
should / be used. The formulas are intended for the deter- 
mination of sections for working, not breaking, loads. 

Still, as already pointed out, there is not a wide disagree- 
ment between experimental results and those given by the 
formula. For the actual strength of iron 
and steel in compression, when used in long 
struts, is little, if any, above the yield point 
(as may be seen from the lov/est curve of 
the diagram, Fig. i). Such struts, if slightly 
bent under a load, fail rapidly when the 
load is increased by a small amount As a 
slight error in centering the experimental 
column has a marked influence on its fibre stress, by the 
introduction of a moment Pj/q , some of the scattered results 
of tests may be attributed to such a cause. 

150. Swelled Columns. — Some posts and struts, especially 
such as are built up of angles connected with lacing, are swelled 
or made of greater depth in the middle. If the strut is perfectly 
free to turn at the ends, such increase in the value of /" may be 
quite effective, and r^ for the middle section may be used in 
determining the value of P, provided the latter does not too 
closely approach the uniform compression value at the narrower 
ends. But if the strut is fixed at the ends, or is attached by a pin 
whose diameter is large enough to make a considerable friction- 
moment, such enlargement at the middle is useless; for an equally 
large value of r'^ ought to be found at the ends also. Hence 
swelled columns and struts are but little used. 

151. Column Eccentrically Loaded. —When the load 
is applied eccentrically to a long column, the maximum unit 




COLUMNS. 149 

stress found in the extreme fibre on the concave side must be 
due to three combined effects: — 

1st. The stress due to the load P, or /o = P ^ S. Fig. 61. 

2d. The stress due to the resisting moment set up by Pj/q . 

3d. The stress due to the resisting moment set up by Vv^ . 

From § 137,/= |(i +-^-^] =, ^^ 4. ^^ .1W^_ 

From § 144, / = — ( I -f a~\ = p^-\- p^a—. 
o r'J 7" 

If then the column is long and the line of action of 
P deviates from the original axis of the column by a distance 
Jo , the three expressions, p^ , /oi'0/1 ~^ ^"^ ^^^ Po ^^' ^ ^^ 
should be added, giving 



/=A(i+ V+-^) 



P = 



^'- , yoy:\ . -n _ /s 



7' - ■ ;- 

Since y^ will determine the direction of flexure, r must 
be taken in this case in the direction y^ ; that is, the moment 
of inertia and r must be obtained about the axis through the 
centre of gravit}^ and lying in the plane of the section, per- 
pendicular to Jo . 

_ That the moment Pjo , although small, has a decided 
weakening effect on a column is proved by experiment, and its 
unintended presence may explain some anomalies in tests. 
See "Experiments on Strength of Wrought Iron Struts," by 
James Christie, Trans. Am. Soc. C. E., Vol. XIII., 
April, 1884. 

Example. — If, in the example of § 148, the load is applied 

36,000 . 7 . 4 r 240 . 240 . 8 3 • 8 ~\ 

Im. off centre, /= i -\ — + I 

0.22V 28,000 .36 2 . 36/ 

= 7,650 (1.8) = 13,800 lbs. The stress on the convex side is 
+ 1.530- 

152. Straight-Line Formula for Columns. — Rankine's 

formula P-^S=/^(i-hrt — ^), when plotted for various 

values of P -^ S = j, and I ^ r — x, gives a rather flat 
reversed curve, as shown in Fig. 62. The first part of the 
curve is not of practical value, as columns v^^hich have small 



I50 



STRUCTURAL MECHANICS. 



F\g 62 



values of / — r can be determined with sufficient accuracy by 
P = fS. The largest values of / ^ r also are of no service, 
as the curve yields too small values of /for use, and ap- 
proaches the quantity P — S 
= 4-^Er^ — /^ The equation 
of a straight line which shall 
nearly coincide with the flattest 
portion of the curve, on either 
side of the point of contra- 
flexure, may be substituted for 
Rankine's formula for conven- 
ience, and the same values of 
mean stress will be practically obtained within working limits 
of / -^ r. 

Rankine's formula, upon the substitution of the symbols 
X and J as above, may be written j = f ^ (^i -|- ax^), and the 
co-ordinates x' and j/' of the points of contraflexure found by 




puttmg 

dy 
dx 



dx"" 



O. Thus 



2afx d'^y — 2<2/"( r -|- ^^^) -\- ^d^fx^ 



(i -|- ax^y dx 



(i -\- ax'^Y 



Dropping factors, — (i -}- ax"^) -j- ^ax"^ = o .-. T^ax"^ — i; 
x' =1 1 ^ s/ '^a, andjF' = f/, the desired co-ordinates. 

The equation of a straight line tangent to the curve at 

the point of contraflexure may now be deduced from the well 

dy' 
known formula j/ — y' = -r-,{x — x'). 



dy' 
dx' 



dx' 



= — i/Vs^- 



y = f/— If \/2>(^{x — 



Vs' 



) = t/— tA/3^ • ^- 



If/ = t6,ooo, E = 29,000,000 and a =■ f — 4-^E, 
P -^- S ^=1 j' ■=. 18,000 — 39 - for fixed ends, and 



/' = 18,000 



78 — for hinged ends. 
r 



('■) 



COLUMNS. 151 

The curve of Fig'. 62 is plotted for these values of/" and 
E, and the tangent at the point of contraflexure is showa for 
a column with fixed ends. 

If y = 14,000, and E and a are as before, there result two 
corresponding values of the mean stress, for fixed and for 
hinged ends. 

/' = 15.750 — 32 — and 15,750 — 64-. (2.) 

;' r 

If it is desired to use a straight line formula for moder- 
ately short columns, such a line might be drawn through 
;r'' z= / -^ r = o and j/" = P -h S = / and through the point 
of contraflexure. Since 

_ V3^ 

y — f— \f ^ z^ ' ^• 

If/"= 16,000, and E and a are as above, 

/' = 16,000 — 26 — for fixed ends, and 

r 

= 16,000 — 52 — for hinged ends. (3.) 

' A straight line through /-^r = o, P^S=: 16,500, and 
the point of contraflexure is shown in the figure, and yields 

/' = 16,500 — 27 — , and 16,500 _ 54 — . (4.) 

If the safe unit stress for live load is taken as one-half its 
value for dead load, the preceding equations (i. ) to (4.) would 
become for live load: — 

Fixed ends. Hinged ends. 

9,000 — 14 — 9,000 — 27 — (i . ) 

;' r 

7,875 — 11— 7,875—22— (2'.) 

8,000 Q 8,000 18 (3'-) 

r r 

8,250 — 10 — 8,250 — 20 — (4'.) 



152 STRX/CTURAL MECHANICS. 

It will be seen that the first term in each case is obtained 
by dividing by two; but the second term must be computed 

from/Vs^^. 

If the value of /for combined dead and live load, when 
they are nearly equal, is taken as 12,000 lbs., there will result 
in place of (3. ) and (3'.), 

12,000 — 17 — and 12,000 — 34 — (5.) 

Such modifications of Rankine's formula are convenient 
for use and are often specified. 

For current formulas see § 175. 

If / = 1,000 and E — 1,400,000, there will be obtained, 
in place of (3.), 1,000 — 2/ -^ r. But, as timber struts are 
usually rectangular, I2r^ = h^, where h is the least dimension, 
and there results 

f = T.ooo — — - for fixed ends. 
-^ Io/^ 

The subtractive term is much smaller than the one com- 
monly employed. See § 170. 

153. Transverse Force on a Column. — The resisting 
power of a column or strut to which a transverse force is 
applied in addition to the load in the direction of the axis, 
and the proper dimensions of the strut, are involved in some 
doubt. Theoretically, the formula is deduced as follows: — 

From the formula for the resisting moment of a beam, 
M — /" I -h jj/j, the stress on the outer layer from such bend- 
ing moment is Mj/j -^ I. Hence, if M is tJiat particular value 
of the bending moment (from the transverse load or force) 
which exists at the section of maximum strut deflection, 
where the colu7nn stress is greatest (that is, at the middle for 
a column with hinged ends, but perhaps at the ends for a 
column with fixed ends, since M may then be greater at the 
ends), the max. unit stress on the concave side 






and 



P = 



(/-T)s /s-^ 



1 






COLUMNS. 153 

when the max. fibre stress does not exceed/. Or, it may be 
said that, at the section in question, P is moved laterally by 
the moment M a distance 7^ = M -4- P. Then by § 151, 

whence P = ^. 

The value of r^ to be used in this formula is that for 
flexure in the plane of M. It may be convenient, when a 
strut of two channels or similar built up members is proposed, 
to determine the depth and section of the channels to ^ive 
the necessary column strengtJi alone, for 7-^ about an axis per- 
pendicular to their webs, and then to place them a sufficient 
distance apart in the plane of the moment M to ^ive a value 
of r^ which will also satisfy the above formula for P. 

Strut-beams are not economical, and their introduction 
should be avoided, if possible. 

, Exa77iple. — A steel column of two 9 in., 2o| lb. channels, 
laced together; combined section 12 sq. in., I for axis perpendicu- 
lar to web = 2 X 58.5, r -=. 2.72, I ^ r =. 90. If the column 
is 20 ft long between pins, its supporting capacity, by (4.) § 152, 
will be P = 12(16,500 — 54 . 90) = 139,700 lbs. 

If a horizontal force of 14,000 lbs. is applied at 4 ft. from 
one end, the distance which these channels should be spaced apart, 
in the plane of the bending moment, if the flanges are turned in, 
is about 20 in. For I of one channel about its own axis parallel 
to web = 2.5; distance of axis from back — 0.56 in.; I for col- 
umn with 20 in. spacing ■=■ 2 [2.5 -f- 6(10 — 0.56)^] = 1,191; 
r^ ^= 99-2; / -^ ;- = 24. P -^ S := 11,640 =■ /' — 1^300; 
f = 12,940 lbs. M at middle of column = ( 14,000 . 4 -^ 20)120 
=z 336,000 in. lbs. /" = 336,000 . 10 -^ 1,191 = 2,820 lbs. 
/'+/" = 15,760 lbs. 

154. Lacing Bars. — The parts of built-up posts are 
usually connected with lacing straps or bars. See XVI., Plate III. 
These bars carry the shear due to the bending moment arising 
from the tendency of the post to bend, and should be able to 
stand the tension and compression induced by the shear. At the 



154 STRUCTURAL MECHANICS. 

ends and where other members are connected, in order to insure 
a distribution of load over both members, batten or connection 
plates at least as long as the transverse distance between rivet rows 
are used in the best practice. 

The value of this shear would be found by taking the first 
derivative of the bending moment, Fv; but v is unknown. As the 
lacing bars are usually of uniform size throughout the length of 
the post, it will suffice to determine their dimensions at the ends 
of the strut. 

From the formula/ = —I i + '^ ^ I it appears that the max. 

P /'^ 
unit stress due to tendency to bend is -— - <^ — „, which may be 

equated with M}\ -^- I. It may be assumed that the moments at 
the several points along the axis of a strut with fixed ends vary as 
do those of a beam loaded with W at the middle and fixed at the 
ends. In the latter case, by § 117, M = JW/ at the two ends and 
at the middle, and the points of contraflexure are at J/ from either 
end. Hence the two curves must be much alike. Then 

P /■' W/y, ■ ,,, 8 Fa/ 
S r^ 8 1 Ji 

As the maximum shear in the above case is g-W, 

4 Fal 



F max. = -^W = 



Ji 



If the strut has hinged ends, replace a by 4a or 2a, as ex- 
plained in § 145. 

Example. — The 9 inch column of the last section, if loaded 
with 140,000 lbs. and spaced 9 in., so that the lacing bars are 8 in. 
long between rivets, will have a =. 16,500 -^ (40 . 29,000,000) = 
I -^ 70,000 nearly. Then F = 4.4. 140,000 . 240 -^ (4^ . 
70,000) =1 1,700 lbs. Pull or thrust in bar is about 2,000 lbs. 
The shearing or bearing value of one f rivet in i in. plate is 2,300' 
lbs. A 2 in. X 4 in. bar in tension, net section /g sq. in. carrying 
2,000 lbs. will have a unit stress of at least 6,400 lbs. In com- 
pression it can carry some 5,000 lbs. 

If this column resists the horizontal force of 14,000 lbs. also, 
the shear will be 770 -{- 2,800 = 3,570 lbs. at one end and 
12,000 lbs. at the other. The bars will be some 21 in. long, and 
must be crossed and riveted at intersections. A continuous plate 
may be advisable where the shear is heavy. 

Each piece should be of equal strength throughout all its 
details. A post or strut composed of two channels, con- 
nected by lacing bars and tie plates, is proportioned for a 



COLUMNS. 155 

certain load, the mean unit stress being reduced in accordance 
with the formula in which the variable is the ratio of the 
length to the least radius of gyration of the whole section. In 
the lengths between the lacing bars, this ratio for one channel 
with its own least radius should not be greater than for the 
entire post. Nor should the flange of the channel have any 
greater tendency to buckle than should one channel by itself. 
The same thing applies to the ends of posts, where flanges are 
sometimes cut away to admit other members. Quite a large 
bending moment may be thrown on such ends, when the 
plane of a lateral system of bracing does not pass through the 
pins or points of connection of the main trus^ system. 

A recent specification reads: Single lacing bars shall have 
a thickness of not less than ^V, and double bars, connected by 
a rivet at their intersection, of not less than eV of the distance 
between the rivets connecting them to the member. Their 
width shall be — For i 5 in. channels, or built sections with 
3^ and 4 in. angles, 2J inches, with J in. rivets; for 12 or 10 in. 
channels, or built sections with 3 in. angles, 2\ inches, with 
I in. rivets; for 9 or 8 in. channels, or built sections with 
2\ in. angles, 2 inches, with | in. rivets. 

The distance between connections of the lacing bars shall 
not exceed eight times the least width of the segments con- 
nected. 

All segments of compression members, connected by 
lacing only, shall have ties or batten plates placed as near the 
ends as practicable. These plates shall have a length of not 
less than the greatest depth or width of the member and shall 
have a thickness of not less than h of the distance between 
the rivets connecting them to the compression member. 

Examples. — i. A single angle -iron, 6 X 4 X f in. and 6 ft. 
8 in. long, is in compression. Use r = 0.9 or obtain it from § 99, 
X. S = 3.61 sq. in. If P -4- S = 12,000 — 34/ -^ r, what will it 
carry? 32,400 lbs. 

2. A square post 16 ft. long is expected to support 80,000 lbs. 
If y = 1,000 and the subtractive term is 2/ -^ r, what is the size? 

10 X 10 in. 

3. A cylindrical rod of steel is 4 in. in diam. and 100 in. 
between pins. What is the max. allowable thrust ii/ = 8,000 lbs. 
for an alternating load and (3') is used? 78,000 lbs. 



156 STRUCTURAL MECHANICS. 

4. What is the safe load on a hollow cylindrical cast-iron 
column 13 ft. 6 in. long, 6 in. external diam. and i in. thick, if it 
has a broad, flat base, but is not restrained at its upper end? 
/= 9,000 lbs., E — 17,000,000, S = 15.7 sq. in. 

124,000 lbs. 

5. If a short wooden post, 12 in. square, carries 28,800 lbs. 
load, and the centre of pressure is 3 in. perpendicularly from the 
middle of one edge, what will be the max. and the mean unit 
pressure, and the max. unit tension, if any? 

500 lbs.; 200 lbs.; — 100 lbs. 



CHAPTER X. 

SAFE WORKING STRESSES. 

155. Endurance of Metals Under Stress. — It is import- 
tant to determine how long a piece maybe expected to endure 
stress, when constant, when repeated, or when varied and per- 
haps reversed; and it is still more important to find what 
working stresses may be allowed upon a given material in 
order that rupture by the stresses may be postponed indef- 
initely. 

The experiments carried on by Wohler^ and Spangen- 
berg, and afterwards continued by Bauschinger, show the 
action of iron and steel under repeated stress. 

156. Woehler's Experiments. — A number of bars of 
wrought iron and steel were subjected to a load which 
repeatedly varied between zero and a certain quantity. One 
series of tests was by direct tension, another series by trans- 
verse bending. When the bar broke under the load, a similar 
bar was tested under a reduced load, and was found to resist 
a greater number of applications before fracture. Under a 
smaller load the number of applications necessary to produce 
fracture was found to increase. Finally a load was reached 
which did not cause failure after some 40 million repetitions. 
This last value was taken as the limiting safe strength for 
loads applied in that particular way. Thus bars of unhard- 
ened Krupp spring steel under a steady load broke with 
110,000 lbs. tension per sq. in.; with 100,000 lbs., fracture 
ensued after 40,000 repetitions; with 90,000 lbs. after 72,000 
repetitions; 80,000, 70,000 and 60,000 lbs. required 132,000, 
197,000 and 468,000 repetitions; and for 50,000 lbs. the bar 

*VVohler, Uber die Festigkeitsversuche mit Eisen and Stahl, Berlin, 1S70.- 
Zeitschrift flir Bauwesen, Vols. X, XIII., XVI., XX. 
See also Burr's Elasticity and Resistance of Materials. 
Technic, Univ. of Mich., i8Sg. 



158 STRUCTURAL MECHANICS. 

endured 40 million repetitions of load from zero without 
fracture. 

In other experiments the stresses were varied between 
different limits and from tension to compression. 

For specimens taken from a certain iron axle, experi- 
ments showed that alternations of stress between the following 
limits of tensile ( — ) and compressive (+) stress per square 
inch, might take place with equal security against rupture with 
several million repetitions. 

— 15,500 lbs. and + 15,500 lbs. Difference 3 1,000 lbs. 

— 29,000 " o " 29,000 '' 

— 43,000 " - — 23, 500 lbs. '' 19,500 " 

it being of course assumed that in all cases the maximum stress 
is less than that required to produce rupture under a static 
load. 

Also unhardened spring steel gave the limits 

— 48, 500 lbs. and o Difference 48, 500 lbs. 

— 68,000 '* — 24,000 lbs. •' 44,000 " 

— 78,000 '* — 39,000 " '* 39,000 '' 

— 87,500 " — 58,500 " ♦' 29,000 '' 

For specimens from a Krupp cast-steel axle the following 
values gave equal security against rupture. 

— 27,000 lbs. and -\- 27,000 lbs. Difference 54, 000 lbs. 

— 47,000 " o " " 47,000 " 

— 78,000 " — 34,000 " " 44,000 " 

and for shearing resistance of a cast steel axle, first in one 
direction and then in the opposite direction, 

— 21,500 lbs. and + 21,500 lbs. Difference 43,000 lbs. 

— 37,000 " o " 37,000 " 

157. Woehler's Laws. — Wohler's laws, deduced from 
many and long-continued experiments, are 

Rupture of material may be caused by repeated alter- 
nations of stress, none of which attains the absolute breaking 
limit. 

Differences of stress, the extreme range of stress, are a 
sufficient cause of rupture; and the absolute magnitude of ex- 



SAFE WORKING STRESSES. 159 

treme stress is controlled by the fact that, as the stress 
increases, the differences which are sufficient to cause rupture 
become less. 

The above examples indicate this law. 

158. Bauschinger's Experiments. — In endurance tests 
made by Professor Bauschinger specimens were repeatedly 
pulled from zero to certain stresses. Many of the specimens 
were several times examined for elastic limit during the pro- 
gress of the endurance tests. The stresses to which bars were 
subjected were varied also in kind and amount, and the num- 
ber of applications required to produce fracture, when fracture 
occurred, were recorded. It was found that, in all cases, the 
number of repetitions of loading borne by th'e bar diminished 
with increased range of variation of stress, and the diminution 
was surprisingly regular. 

He derived the following values of the highest tensile stress 
which may be applied to each material indefinitely, beginning 
each pull at zero, or what may be termed the maximum safe 
tensile sti'ess, for working stress from zero. There is also 
given the maximum permissible range of stress, when the 
extremes are of opposite kinds, and when they are of the same 
kind. See p. i6o. 

To these values are added the elastic limit, yield point, 
and breaking stress for a single static application, all in pounds 
per square inch. Values for different samples will not agree 
exactly, and must be considered as approximate averages. 

If a piece has applied to it a tensile stress which reaches 
the yield point, it can, after a period of rest, have applied to 
it with safety an equal stress in compression. When, how- 
ever, the stresses repeatedly alternate between plus and minus, 
the algebraic difference between them, the numerical sum must 
be taken into account, and the evidence seems to be that such 
sum cannot, for continued repetitions of stress, ever safely 
exceed the distance between zero and the yield point. 

To repeat a previous statement: — Elastic limit marked 
that load at which either permanent set could first be detected, 
or where increments of stress and extension ceased to be 
proportional to each other, the earliest certain and contin- 
ued indication either way being taken to mark that limit. 



i6o 



STRUCTURAL MECHANICS. 



The first indication of permanent set was very strictly noted 
as the elastic Hmit. Yield point marked that load at which 
rapid and considerable yield first took place under a steady 
load, and was the point called elastic limit in ordinary testing, 
done without very accurate means for measuring small exten- 
sions. The original elastic limit was far below yield point in 
many of the specimens, notably so in the iron plate, where 
the difference is fifty per cent, of the yield point, and in the 
steel plate where the difference is twenty per cent. 

bauschinger's endurance tests. 

Stresses requiring five to ten million repetitions to cause 
fracture, in pounds per square inch. 



MATERIAL, 



Wrought Iron 

Bar Iron 

Mild Steel Plate 

Steel Axle 

Steel Rail 

Mild Steel Boiler Plate 



OPPOSITE 


STRESSES. 


ONE STRESS 
ZERO. 


SIMILAR S 


Least. 


Greatest. 




Greatest. 


Least. 


— 16,000 


+ 16,000 





29,000 


25,500 


— 17,500 


-(- 17,500 





32,000 


30,000 


— 19,000 


+ 19,000 





35,000 


29,000 


— 19,000 


+ 19,000 





35.000 


32,000 


— 23,000 


+ 23,000 





43,000 


44,000 


— 21,000 


+ 21,000 





43,000 


43,000 


— 19,000 


+ 19,000 





35,000 


29,500 



Greatest. 

43,000 
49,000 
49,000 
53,000 
71,000 
69,000 
50,500 



MATERIAL. 


ELASTIC 

LIMIT. 


YIELD POINT. 


ULTIMATE 
STATIC. 


ELONGATION 
PER CENT. 


Wrought Iron Plate 


15,000 
26,000 
32,000 
34,000 
38,000 
40,000 
37,000 


30,000 
32,500 
35,000 
42,000 
47,000 
43.500 
42,000 


51,000 
57.500 
57,000 
62,000 
87,000 
85,000 
58,500 


15-5 
12.5 
14. 


Bar Iron -^ 

Bessemer Mild Steel Plate 


Thomas Steel h xle 


18.5 
195 
26. 


Thomas Steel Rail 

Thomas Steel Boiler Plate 



The steel was somewhat irregular. 

159, Maximum Tensile Stress. — The maximum safe 
tensile stresses from zero, "regarded as more or less rough 
approximations, show that every such stress is near and below 
the yield point. Hence the limiting safe stress on sound bars 
of iron and steel, for tensions alternating from zero to that 
stress, is just a little less than the original yield point of the 
metal. 



SAFE WORKING STRESSES. l6l 

i6o. Elastic Limit and Alternating Stresses, — It ap- 
pears that a bar subjected to alternate pressure and tension 
will break after a sufficient number of repetitions with a stress 
less than its primitive elastic limit. But it was found that the 
application of a stress exceeding the elastic limit raised that 
limit, — in certain cases, nearly to the breaking stress. Prof. 
Bauschinger, in 1886, said that tension greater than the true 
elastic limit applied to a bar raised its elastic limit in tension, 
but not without lowering the limit in compression, and vice 
versa. Even a moderate raising of the tension limit may 
lower the compression limit to zero. Therefore the law that 
the elastic limit can be raised by stress does not apply to 
alternating stresses. Further, these artificially produced elastic 
limits are extremely unstable. He advanced the view that the 
primitive elastic limit of many materials is an artificially raised 
one. The material has been subjected to mechanical opera- 
tions in manufacture which are equivalent to straining actions. 
Alternate compression and extension has the effect of raising 
an artificially lowered, or lowering an artificially raised, elastic 
limit. By subjecting a bar to a few alternations of equal 
stresses, which are equal to or somewhat exceed the elastic 
limits, the latter tend towards fixed positions, which Bausch- 
inger called the natural elastic limits. The range of stress 
for which a bar is perfectly elastic after a few repetitions of 
such alternating stresses appears to agree very closely with- 
Wshler's range of stress for unlimited repetitions of alternating 
stresses. 

A curious discovery, that moderate hammering on the 
end of cast-iron bars appears to relieve internal stresses, per- 
haps cooling stresses, and to increase the tested strength,, 
seems to point in the same direction. 

161. Alteration of Structure. — In discussing the ques- 
tion of internal changes, Unwin says:^ It would appear 
likely that any gradually progressive alteration or fatigue of 
the bar would be manifested in some way in alteration of the 
strength, the elastic limit, or the elongation of the bar when 
tested in the ordinary way. But this, so far, appears not to 

*The Testing of Materials of Construction: Unwin. 



l62 STRUCTURAL MECHANICS. 

be the case. A bar subjected to so many repetitions of load- 
ing that it is known to be on the point of breaking, or a piece 
of a bar already broken in an endurance test, gives in the 
testing machine no indications that the strength or ductility 
has been altered. See §i66, (/. ). It is in accordance with 
experiments on pieces of structures long subjected to loading; 
and no one would guess that these test pieces were in any 
respect different from new material. But the fact still remains 
that material subjected to repeated loading is different from 
new material. The material, after a certain number of repe- 
titions with a given range of stress, does break with fewer 
subsequent repetitions. 

Whatever the alteration produced by repetition may be, 
it certainly does not appear to be a loss of strength (statical 
resistance). If it is a loss of ductility or power of elongation, 
it must be a loss confined to very short portions, or planes of 
weakness, in the bar; for, if not, it would be shown in ordi- 
nary testing. In certain cases flaws or fissures have been 
found to be present in bars subjected to so many repetitions 
of load that they were on the point of breaking. It is at least 
conceivable that repetition of stress picks out sections of 
weakness in the bar, and that the deterioration is confined 
almost to such planes. The deterioration may be primarily a 
loss of power of yielding in the particles near the plane of 
weakness, and not a loss of tenacity. Such a loss of ductility 
at a section might well show itself finally in a rapidly spread- 
ing fissure or crack. This explanation is purely hypothetical, 
but it is at least in accordance with a curious fact. Bars frac- 
tured in the Wohler machines usually showed no trace of 
drawing out, however ductile the material might be, tested 
statically. They broke as if the material had been perfectly 
brittle. This peculiar fracture, without indication of any 
plastic drawing-out, is not . uncommon in fractures of tires, 
axles and other structures in ordinary practice. 

162. Launhardt-Weyrauch Formula. — Two formulas 
have been advanced, based on Wohler's experiments, for the 
determination of the allowable unit stress of either sign on 
any material when the range as well as the magnitude of the 
stress is considered. Launhardt proposed the following 



SAFE WORKING STRESSES. 163 

formula for the breaking load of a member which is subjected 
to (apparent) stresses, varying between maximum and mini- 
mum stress of the same kind: 



(, / — u min. stress \ 
I A I, 
u max. stress./ 



in which a is the breaking weight or stress under the condi- 
tions to which the member is subjected, u is the greatest stress 
which the piece will bear without breaking, if repeated from 
zero to 2L an indefinite number, many million, of times, and t 
is the breaking stress at a single application. 

Weyrauch extended Launhardt's formula, to cover cases 
of alternate tension and compression, in which case min. 
stress is to be considered as negative, giving 

(, u — V min. stress'^ 
I A I, 
u max. stressy 

where v is the greatest stress which the piece will bear with- 
out breaking, if repeated from -\- v X.o — z; an indefinite num- 
ber of times. 

In some of Wohler's experiments il appears to be greater 
than \ /, and to approximate | /. This value is assumed by 
Weyrauch and gives {t — u) ^ ^^ =" h- Also v by Wohler's 
experiments is equal to |^ /, or (?/ — v) -^ 21 =^ J. Therefore 
both formulas become 



f , mm. stress "^ 

= ?/ I I -j- ■ I, 

V 2 max. stress^ 



the sign of the second term changing when the minimum 
stress and maximum stress have opposite signs. 

When a is the maximum stress, as it must be in design- 
ing a member which is to be subjected to minimum stress 
also, 

,T , / N min. stress 

ivlax. stress = u -\- (t — ?/) . , or 

max. stress 

Max. stress^ — zi max. stress -\- ^ 7i^ = ^ li'-' -\- {t — ti) min. stress. 

Max. stress r= i 2^ -[- -j/ (^ ?/^ -|- (/ — u') min. stress), 
or from above, 

,/",,/, 2 min. stress x'\ 
Max. stress ^ \ u \\ -\- y \^ -\- ) \. 



164 



STRUCTURAL MECHANICS. 



The maximum stress or a must then be reduced by a 
reasonable amount to cover the uncertainties enumerated in 
§ 167, so that i (^ or ^ (^ is taken for the maximum safe unit 
stress in designing. It also appears that for steady load or no 
variation, a ^ t\ for a load from zero to ?/, a ^= u =^ \ t\ for 
a load from -\- v to — v, a = v =^ ^ L Experiments show 
more or less disagreement, as is to be expected, but the aver- 
age results are as indicated. 

163. Fidler's Dynamic Theory. — Fidler* suggests the 
following line of reasoning, as a dynamic theory, to account 
for the different appareiit values of the unit stress which a 
material will endure when it is subjected to variable forces 
ranging between known limits. 




c f/ 


/ 

D 


V 




fo 





°< ^ ^ 



> 




As previously explained, a static load Pq causing a unit 
stress B A, Fig. 63, will do work of elongation, (or compres- 
sion) in increasing from zero to B A, equal to the area of the 
triangle O B A. A suddenly applied load Pj which, if static, 
would cause a stress O C, since it does work equal to the pro- 
duct of O C into the elongation, must not be greater than 
one-half the static load, in order that O B = / and B A == / 
max. may not be greater than before; for then area O B D C 
= area O A B. The real stress in this latter case is twice the 
apparent stress found by dividing the load by the cross-section 
of the piece. 

If the piece is already under a steady load stress H E, 
Fig. 64, which has caused the elongation (or compression) 
O H, the sudden imposition in addition of a load which, if 

* Practical Treatise on Bridge Construction, T. C. Fidler; Pliiladelphia, 
Lippincott. 



SAFE WORKING STRESSES. 1 65 

static, would cause the added stress E C, will do work equal 
to the rectangle E D and produce the final elongation O B = )^, 
since triangle E C F := triangle FDA. Then the real stress 
/ max. = /o + 2/1, or = (P^ -|- 2P1) -f- S. Remember that / max. 
and / are directly related. 

Similarly, if /^ = J/ max., and f^ is due to a sudden 
increase of load, /, can only be one-half of /max. — /^ , if 
/ max. and / are to be kept within safe limits. Then, in 
general 

Actual stress = f^ -\- 2/j = (Pq + ^Pj) -^ S, (i.) 

Apparent stress = (Pq + Pj) -^- S. (2.) 

If Pq is of the opposite sign to Pj, Fig. 65, the elonga- 
tion from Pq will be O H and the stress — /^ = H E. The 
work done by Pj, if E C = D A = /, must be a rectangle equal 
toECxCD = E C F + FDA, and/ max. = A D + 
DB = AD + CE — HE. 

Actual stress =2/^ — /^ z=: (2P1 — Pq ) -;- S. 
Apparent stress = (P^ — Pq ) -f- S. 

The former expressions (i.) and (2.) are general, if the 
sign of Pq is contained in the symbol. In the first case, Pq = o. 

The stress will be equal to the load divided by the cross- 
section when the bar is in equilibrium, or at rest under a static 
load. When a part or all of the load is suddenly applied, the 
stress will momentarily be greater, then will be less, the bar 
will be thrown into vibration, and finally come to rest. When 
a load is applied rapidly but not suddenly, the maximum stress 
will be intermediate between that caused by a static and by a 
sudden loading. 

On the above line of reasoning 

Equivalent static load = P^ -|- 2P1 = Pq -h P] + Pi 
= Max. load + (Max. — Min.) load, 

or the maximum load of the strain sheet should be increased 
by the amount of the suddenly applied load, (or the max. 
stress — min. stress), and then used as a static load for the 
determination of the necessary cross-section. Members like 
stringers, cross-girders, and vertical suspenders in bridges, and 



l66 STRUCTURAL MECHANICS. 

short girders should doubtless have this full allowance made. 
Those parts of bridges of considerable span, which do not 
receive full pull or thrust until a large portion of the span is 
loaded may have only one-half of this allowance added. See 
Pencoyd Bridge Co. 's allowance for impact, § 172. 

Structures like roofs, subjected at long and rare intervals 
to the greatest loads for which they are computed, need very 
little if any allowance for variation of stress. 

164. Seefehlner's Rule. — Seefehlner has proposed the fol- 
lowing rule, founded on Wohler's experiments: — 

Let P = total maximum force, and P' = total minimum force 
to which the bar is alternately subjected; n — ratio of P' to P; 
u = rupturing or ultimate unit stress for a single application, as in 
a usual test; / = max. allowable unit stress, approaching the 
dangerous limit. Then 

P' 



-^='^(' +-^)^ = *(^ ± 2^^)^ 



If P' is of the opposite kind to P, their ratio will be negative; 
hence the alternative sign is given to n. 

Examples. — If u z=. 50,000 lbs. per sq. in. and 

P^ = o P = 60,000 lbs.; / =r |(i -(- o) 50,000 = 33,300 lbs. 

P^ = 24,000 lbs., P = 72,000 lbs.; / — |(i -j- \) 50,000 = 38,900 lbs. 
P^ = — 72,000 lbs., P =: 72,000 lbs.; / — |(i — \) 50,000 — 16,700 lbs. 

165. Gerber's Parabola. — Suppose that the ranges of stress 
for unlimited repetition are known for any material and are plotted 
as ordinates and the corresponding minimum stresses as abscissas; 
the points so located will fall in a parabolic curve. 

Let f max. and / min. = limits of stress. 

D = /"max. =F /min. = range of stress, — being used for the 
same kind, and -}- for different kinds. 

f = statical breaking stress. C = constant for the material. 

Then Gerber's equation is 

f :^: (/-min. +iD)^-f CD. . 

If / is known, and also f min. and f max. for any one range of 
stress, for unlimited repetitions before breaking, C can be deter- 
mined, and then the limits of stress for all conditions of loading 
can be calculated. 

For curves, see Unwin, '^Testing of Materials of Construc- 
tion," p. 392. 

166. Bauschinger's Laws. — (^a). The yield point is 
always immediately raised to that load with which a bar has been 
stretched. Thus, if the diagram first obtained is A B D, Fig i, the 



SAFE WORKING STRESSES. 1 67 

next application of the same or greater force will give C D. Dur- 
ing a following period of rest,- moreover, the yield point rises 
above the greatest load with which the bar has been stretched, 
quite noticeably in one day, and continues to rise for weeks and 
months. 

[A bar must then experience some readjustment of particles 
after strain, in the interval of rest.] 

(/;•. ) The true elastic limit (limit of proportionality of change 
of length to stress producing it), is lowered, often to zero, by 
stretching the bar beyond the yield point; so that the specimen, if 
tested again, immediately after stretching and removal of load, 
has no true elastic limit, or a remarkably low one. But in a fol- 
lowing period of rest the elastic limit rises again; it reaches, after 
several days, the load with which the bar was stretched, and, after 
sufficient time, certainly after some years, it is, raised even above 
that load. 

The distinction in behavior between elastic limit and yield 
point is quite wide. Mr. C. A. Marshall* did not find that the 
elastic limit was lowered by the action of a stress exceeding the 
yield point, provided the highest load was allowed to act long 
enough to produce its full extension. 

(^. ) The true elastic limit is raised by pulling a bar with loads 
above the elastic limit but below the yield point, and continues to 
rise after the loadings have ceased, and it rises the more the higher 
the load. The elastic limit reaches a maximum when it has been 
raised to near the yield point, and, by the application of loads 
beyond the yield point, it is thrown back according to law [I?.) 
above. 

-Examples of elastic limits raised above the original yield point, 
as determined from a duplicate specimen, are not wanting, and it 
is the rule that the yield point is considerably and unmistakably 
raised by repeated stresses. Hence — 

(d.) By often repeated tensions, of magnitude greater than 
the elastic limit and less than the yield point, the original yield 
point may be raised. 

{e ) By straining in tension beyond the elastic limit, the limit 
for compression is notably lowered; that is brought towai'ds zero; 
and, conversely, by similar straining in compression, the elastic 
limit for tension is lowered. The higher the stresses, the greater 
the change. When an elastic limit, so lowered, is raised again by 
stresses in like direction and is then overstepped, the elastic limit 
for the opposite stress falls at once to zero or almost to zero. A 
rest of three or four days, and even weeks, has but little, if any, 
influence on these processes. Mr. C. A. Marshall found, however, 
that the elastic limit in compression for small steel rounds was 
completely restored by a rest of eighteen months, after the bars 
had been stretched by a load equal to yield point. The bars 

*Technic i88g, Enghieering Society, University of Michigan. 



1 68 STRUCTURAL MECHANICS. 

showed even a higher elastic limit and yield point than could have 
been found originally, viz., each about 41,000 pounds per square 
inch. 

(/. ) By gradually raised stresses, alternating between tension 
and compression, the elastic limit for the opposite stress cannot be 
lowered until the stresses pass the original elastic limit. 

(g.) When an elastic limit for stress of one sign has been 
lowered by a previous straining of the opposite sign, it can be 
raised again by gradually increasing stresses alternating between 
tension and compression, but only to a limit which lies consider- 
ably below the original elastic limit. The limit which it is possible 
thus to reach is thought to be the limiting magnitude of equal and 
opposite loads which may be borne with safety. [May it not be sup- 
posed that the material has by the treatment been relieved of the 
stresses set up in the process of manufacture and brought to a 
normal state?] A member will not be broken by repeated reverse 
stresses until it is loaded up to and beyond these new limits. 

(//. ) Rupture is not caused by from five millions to sixteen 
millions of repetitions of tensile stresses whose lower limit is zero and 
whose upper limit lies in the neighborhood of the true elastic limit. 

(/. ) By often repeated stresses between zero and an upper 
stress which lies near to, or more or less above, the original elastic 
limit, this limit is raised beyond, often far beyond, the upper limit 
of the stresses; and it is raised the higher, the greater the number 
of repetitions, without, however, the possibility of exceeding a 
certain height. 

[k.^ Repeated stresses between zero and an upper limit, such 
as may raise the original elastic limit above the higher stress, do 
not produce rupture. If, however, the upper limit lies so high that 
the elastic limit cannot be raised beyond it, then rupture must fol- 
low after a limited number of such stresses. 

This statement is more general than the earlier one, that the 
limiting safe stress lies just below the yield point. Although the 
yield point may rise during an endurance test, and the elastic limit 
may rise even above the original yield point, yet if the load at the 
beginning of a test exceeds the origi7ial yi^ld. point, and is repeated 
continuously, rupture will finally occur. It however appears to be 
possible to begin with a rightly chosen load, and to gradually 
increase it as the repetitions proceed, keeping at all times below 
the rising yield point, without producing rupture. In most prac- 
tical cases such treatment cannot be applied. The yield point is 
easily determined for most specimens of iron and steel, while the 
trial to raise the elastic limit is slow, difficult and expensive. 

(/. ) The ultimate tensile strength is not lessened by millions 
of repetitions of stress, but is, if anything, raised, when the test 
piece, after such repetitions, is broken by a quiet load. 

(;;z.) Stresses repeated millions of times on iron and steel 
cause no alteration of structure. 



SAFE WORKING STRESSES. 1 69 

The alteration of structure which is found at the surface of 
rupture of a bar broken by repeated stresses appears to be confined 
to the fractured surface itself, as has been directly proved by etch- 
ing the fractured surfaces. At the minutest distance below the sur- 
face of fracture the original structure is again evident. Hence the 
above statement. 

There is, however, a peculiar marking which appears on the 
surface of a polished specimen while being stretched beyond the 
yield point, which amounts to an indication of final failure, and 
shows that the specimen has passed the yield point. 

This evidence of no alteration of structure confines the idea of 
the '' fatigue of metals " to cases in which complete recovery from 
stretch never takes place upon removal of stress. Instead of metal 
being fatigued, or being brought nearer to rupture by working 
stresses well below the limit given in [g. ), it is really improved in 
tenacity and elasticity, while the ductility is not notably decreased 
nnless there are flaws. 

Fractures by repeated loadings below the apparent yield point 
of the metal as a whole are always detail breaks. Continuity is 
destroyed first at a flaw, or an overstrained spot, and from that 
point the fracture spreads by very minute, successive encroach- 
ments, until it has so weakened the cross-section that large 
inequalities of stress, amounting to the same thing as a cross- 
breaking stress, § 137, are produced in the yet sound portion, 
and the piece breaks with a fracture of the remainder resembling 
that produced by nicking and breaking by bending. 

When there is time for the metal to rest, the piece assumes its 
former state. If it be strained up to the elastic limit in tension 
and then be allowed to rest, it can be strained safely afterwards to 
the elastic limit in compression. When, however, the stresses 
repeatedly alternate between compression and tension, plus and 
minus, the algeb7'aic difference must be taken in determining the 
maximum allowable or limiting stress, which is the same as taking 
the numerical sum of the thrust and pull, or the range of stress, to 
guide in determining the safe stress or necessary section. Experi- 
mental evidence seems to show that, for continued repetitions of 
stress, such sum or range cannot ever exceed with safety the differ- 
ence between zero and the elastic limit in either direction. Pe- 
riods of rest between the intermediate applications of reverse stresses 
change the rule and permit of higher stresses. So some structures 
and some pieces will endure safely higher stresses than others. 
The beginner should fix in his mind the fact that pieces and parts 
frequently subjected to rapid change of stress in magnitude and 
sign, must be exposed to far lower unit stresses, if the life of the 
piece is of consequence. 

167. Reduction of Unit Stresses. — The safe working 
stress to be used for any material will depend upon several 
considerations: — Whether the structure is to be temporary or 



lyo STRUCTURAL .MECHANICS. 

permanent; whether the load is stationary or variable and 
moving; if moving, whether its application is accompanied by 
severe dynamic shock and perhaps pounding; whether the load 
assumed for calculation is the absolute maximum; whether 
such maximum is applied rarely or likely to occur frequently; 
whether the stresses obtained are exact or approximate; 
whether there are likely to be secondary stresses due to 
moments arising from changes of the assumed frame; what the 
importance of the piece is in the structure, and the possible dam- 
age that might be caused by its failure. 

The allowable unit stresses of different kinds, and for 
greater or less change, of load, will be further reduced to pro- 
vide against: — Distribution of stress on any cross-section some- 
what different from that assumed; variations in quality of 
material; imperfections of workmanship, causing unequal dis- 
tribution of stress; scantness of dimension; corrosion, wear or 
other deterioration from lapse of time or neglect; lack of 
exactness of calculation. 

The allowable unit stresses so determined will be but a 
small fraction of the ultimate or breaking strength of the 
material; and it is evident that the idea that it Vv'ill require 
several times the allowable maximum working load to cause a 
structure to fail is seriously in error. 

Over-confidence of the inexperienced designer in the cor- 
rectness of his design may be checked by a study of this 
section. 

i68. Load and Impact. — The design should be com- 
pletely carried out, both in the principal parts and in the 
details. The latter require the most careful study, that they 
may be at once effective, simple and practical.* All the 
exterior forces which may possibly act upon a structure should 
be considered, and due provision should be made for resisting 
them. The static load, the-live load, pressure from wind and 
snow, vibration, shock and centrifugal force should be pro- 
vided for, as should also deterioration from time, neglect or 
probable abuse. A truss over a machine shop may at some 
time be used for supporting a heavy weight by block and 

* A portion of these paragraphs is extracted from a lecture by Mr. C. C. 
Schneider. 



SAFE WORKING STRESSES. 171 

tackle, or a line of shafting may be added; a small surplus of 
material in the original design will then prove of value. 
Lio^ht, slender members in a bridge truss, while theoretically 
able to resist the load shown by the strain sheet, are of small 
value in time of accident. The tendency from year to year is 
towards heavier construction. 

Secondary stresses, as they are called, are due — first, to 
the moments at intersections or joints, when the axes of the 
members coming together at a connection do not intersect at 
a common point; and second, to the moments set up at joints 
bv the resistance to rotation experienced by the several parts 
when the frame or truss is deflected by a -moving load. If 
symmetrical sections are used for the members, if the con- 
necting rivets are symmetrically placed, and if the axes of the 
intersecting members meet at one point, secondary stresses 
will be much reduced. 

All menibers of a structure should be of equal strength, 
and the connections should develop the full strength of the 
body of the members connected. The connections should be 
as direct as possible. When a live load is joined to a static 
load, the judgment of the designer, or of the one who prepares 
the specifications for the designer, must be exercised. A 
warehouse floor, to be loaded with a certain class of goods, 
has maximum stresses from a static load. The floor of a 
drill-room, ball-room or highway bridge receives maximum 
loading from a crowd of people, the possible density of which 
can be ascertained. But if these masses of people keep 
moving, and more particularly if they keep step, the effect of 
their weight will be increased by the vibrations resulting there- 
from. This action is generally called impact. 

In the case of a building, the floor-joists, receiving the 
impact directly, will be most affected; the girders which carry 
the joists will be less affected; and the columns which support 
the girders will receive a smaller percentage of the impact, the 
proportionate effect growing less as the number of stories 
below .the given floor increases. In the absence of trust- 
worthy data from which to determine this impact, it is left to 
the judgment of the engineer to increase the live load by a 
certain percentage, or to decrease the allowable unit stress, 



172 



STRUCTURAL MECHANICS. 



for each case, to provide for the effect, as will be seen in the 
values given later. 

The stresses produced by impact in railroad bridges are 
even more uncertain and ambiguous. The assumption is 
made by some that the effect of impact upon the several mem- 
bers is dependent upon the length of loaded distance which 
produces the maximum live load stress in a member, and a 
•certain percentage in accordance therewith is added to the 
live load stresses. Others use different allowable working 
stresses on different members, depending upon the rapidity 
with which the load may be applied, and the proportion which 
the live load on the member bears to its static load. 

For economy, make designs which will simplify the shop- 
work, reduce the cost and ensure ease of fitting and erection. 
Avoid an excess of blacksmith work and much use of bent 
pieces. 

169. ■ Working Stresses for Timber. — Average safe unit 
stresses, in pounds per square inch, for the more common 
woods, when subjected to moving loads with considerable 
shock. The last five lines have sometimes been prescribed 
for railway bridges. For static loads add 50 per cent. 



White oak 

Southern long leaf or Georgia pine 

Douglas or Oregon fir or pine 

Northern or short leaf yellow pine 

Norway pine 

Spruce 

VVhite pine 

California redwood 

White Oak , 

Long leaf Southern pine , 

Oregon pine or fir , 

Spruce 

White pine 



TENSION. 



1,100 

1,000 

1,400 

700 

600 



200 
60 



50 



1,000 
1,200 
1,200 

goo 

800 

800 50 

70o| 50 
700 



COMPRES- 




SION. 


















ca 


5 
< 


6 

Q 






c/; 


S VI 


X 


c 


S£ 


.— < 


ei 




^ 


u 


X '^ 


^ 


< 


a 


1,400 


500 


1,000 


1,600 


350 


1,200 


1,600 


300 


1,100 


1,200 


250 


1,000 


I 200 


200 


700 


1,200 


200 


700 


1,100 


200 


700 
800 

1,200 


1,000 


300 


1,000 


250 


1,200 


900 


I go 


1,200 


800 


150 


800 


Soo 


150 


1,000 



200 

150 

150 

100 



100 
100 



^35 
100 
150 

85 
85 



SAFE WORKING STRESSES. 173. 

In order to be on the safe side, low working stresses are 
usually assumed. If the actual maximum unit stress which 
could possibly come upon a member could be determined, 
including the secondary stress produced by the deformation of 
the system, a unit stress of considerably greater intensity 
might be used. 

170. Working Column Formulas for Timber. — Form- 
ulas for wooden columns and posts have remained without 
much change for a number of years. The most common 
formula for the mean unit stress, P -:- S, is, for the ends 

-1^1 . 1,000 . 1,000 . 1,000 
Flat, Y2 — ^ ^^^ Pi^' n — ^ ^"^o pms, 



I + — --ri. I + — -T2 ' I + 



250 Jr 190 /^^ 125 h^ 

in which / = length in inches between bearings, and h =: 
width of member in inches in the direction of greatest liability 
to bend. As most timber struts are rectangular, h is more 
convenient for use than r, the radius of gyration. For dif- 
ferent woods, insert in place of i,ooo the compression value 
from the preceding table. 

For roofs, if liberal allowance has been made for snow 
and wind pressure, these stresses may be increased 50 per cent. 

For railway bridges and structures exposed to similar 
loading, when some allowance for deterioration from exposure 
is desired, the following formulas may be used: — For yellow or 
Southern pine, 

/ / / 

Flat, 860 — 1 ~r '^ ^^^ P^^' ^^° — ^ T~ ^ T^no pins, 860 — 9 — ;. 

ft fC /I, 

and for white pine, 

Flat, 540 — 6 — ; One pin, 540 — 6 J — ; Two pins, 540 — 7 — .. 

For highway bridges the above column formulas may be- 
increased 25 per cent. ; for roofs, 50 per cent., if liberal allow- 
ance for loads has been made. 

171. Former Tension Specifications for Iron and Steel. 
Permissible tensile stresses have gradually been modified. 
At first little or no discrimination was made between the effects- 



174 



STRUCTURAL MECHANICS. 



of dead and live load, and a unit stress of 10,000 lbs. per 
square inch was generally used. Later, the allowable unit 
stress was modified for different parts of a structure, as shown 
by the following allowed tensile stresses for railway bridges in 
pounds per square inch: — 

On lateral bracing, 15,000; on bottom chords and main diag- 
onals (forged eye bars), 10,000; on do. do. (plates and shapes), 
net section, on counter rods and long verticals (forged eye bars), 
on solid rolled beams used as cross floor beams and stringers, on 
bottom flange of riveted cross-girders and riveted longitudinal 
plate girders over 20 feet long (net section), 8,000; on bottom 
flange of riveted longitudinal plate girders under 20 feet long (net 
section), 7,000; on counter rods and long verticals (plates and 
shapes), net section, 6,500; on floor beam hangers, and other 
similar members liable to sudden loading (bar iron with forged 
ends), 6,000; do. do. (plates and shapes), net section, 5,000. 

For spans exceeding 150 feet, the above allowed tension in 

bottom chords and main diagonals and the compression in top 

chord sections, § 174, might be increased by 
i 

/' 150 X stress from dead load 

Vstress from dead and live load 



— 50 I per cent. 



The effect of frequency and rapidity of loading, the ratio of 
live load stress to dead load stress, and the difference between 
bar iron and shape iron, as well as possibly the more even distri- 
bution of stress in a bar than in a rolled shape, caused the variation 
in values for unit stresses. Steel was not then in use for bridges. 

Waddell, in 1887, gave, for highwa}^ bridges, loaded to 
100 pounds per square foot, and spans up to 150 feet, unit 
stresses in pounds per square inch: 



IRON. 
10,000 



8,000 



7>500 



10,000 



Lower chord bars and end main diagonals (forged eye bars) 

Lower cliords (plates and shapes), net section 

Middle panel diagonals and counters, adjustable 

Hip verticals (forged eye bars) 

Middle panel diagonals and counters (plates and shapes), net 

section 

Hip verticals (plates and shapes), net section 

Flanges of rolled beams 

Flanges of built beams (net section) 

Beam hangers, loops ,... 7,000 

Beam hangers, plates 6,000 

Lateral and vibration rods 15,000 

Values to be increased with increase of span. 

Two other specifications, based on the range of stress 
from maximum to minimum, are cited in § 174, in connection 
with the specifications for compressive stress. 



STEEL. 
12,500 

11,000 

10,000 

14,000 

9,000 



SAFE WORKING STRESSES. 175 

The specifications for railway bridges first referred to were 
somewhat modified later. They gave the following allowable ten- 
sile stresses on wrought iron, in pounds per square inch, and per- 
mitted the use of steel as below: — 

Floor-beam hangers and similar members liable to sudden 
loading (plates or shapes), net section, 5,000; do. do. (bar iron 
with forged ends), 6,000; lateral bracing, 15,000; solid rolled 
beams, used as cross floor-beams and stringers, 8,000; bottom 
flanges of riveted cross-girders, net section, 8,000; bottom flanges 
of riveted longitudinal plate girders, over 20 ft. long, used as track 
stringers, net section, 8,000; do. do. under 20 ft. long, net section, 
7,000; bottom chords, main diagonals, counters and long verticals 
(forged eyebars), 8,000 for live loads, 16,000 for dead loads; 
do. do. (plates and shapes'), net section, 7,500 for live loads, 
15,000 for dead loads. 

Medium steel might be used for tensiofi members, plate 
girders and rolled beams with 20 per cent, increase of above 
stresses, and soft steel in place of wrought iron for all riveted 
work, under certain shop restrictions. 

Union Pacific Railway bridge specifications, 1895, read: — 

For live load, iron in rolled bars and in flanges of girders, 
10,000 lbs. per sq. inch; plates and shapes, 8,000 lbs. 

Unit stresses for dead load, to be allowed at fifty per cent. 
greater than the above; or, add | dead load stress to live 
load stress and use the sum as all live load in computing 
sections. 

For steel of 62,000 lbs. minimum strength, unit stresses 
I 5 per cent, greater than the above are allowed. 

172. Impact. — Pencoyd Iron Works, railway bridge spe- 
cifications. — The effect of impact and vibration shall be consid- 
ered and added to the maximum stresses resulting from engine 
and train-loads. The effect of impact is to be determined by 
the following formula: — Impact to be added to live load stress 

= calculated live-load stress ( — |; where L = leno^th 

VL + 3007 

of loaded distance in feet which produces the maximum stress 
in member. The unit stress then allowed per square inch 
(/. e., reduced to a static basis) is on soft steel, 15,000 lbs.; 
on medium do., 17,000 lbs. 

Prof. Melan proposes the following formula for percent- 
age of increase of live load on railway bridges to bring it to 
an equivalent static load. If L is the span in feet. 



176 



STRUCTURAL MECHANICS. 



Percentage of increase = 14 + 2,600 -^ (L -[- 33), 
which reduces practically to 33^% for L = 100, and 25% for 
L = 200. 

173. ^A^orking Stresses: Tension. — The following re- 
quirements are acceptable for railway bridges: — 



Wrought Iron. 

min. stress 



l-7.50of I + 



max. stress / 
stress~\ 
max. stressy 
5.300 



Soft Steel. 
8,600 (do.) 

7,700 (do.) 

6,000 
23,000 
11,500 
17,000 



Medium 

Steel. 

9,400 (do.) 

8,400 (do.) 

6,600 
25,000 
12,500 
18,500 



Iron. 


Soft Steel. 


Medium Steel 


15,000 


18,000 


21,000 


12,000 


14,000 


16,000 


20,000 


23,000 


25,000 


15,000 


18,000 


21,000 



Chords, ties, counters, and } 

long suspenders. 
Plates and shapes, and bot- ) /- 

torn flanges of built gird- > 6, 70o| i -j- 

ers, ) V 

Hangers, through pinhole, 

Lateral and cross-section ) 

J.J >• 20,000 

rods, wind, ) 

Do., centrifugal force, • 10,000 

Long'l rods in trestles, 15,000 

For highway bridges, add 25^. 
For roofs and iron buildings: — 

Bars, main members. 

Shapes, net section, and bottom 

flanges of rolled beams. 
Lateral bars. 

Lateral angles, net section, 

174. Accepted Column Formulas. — Column formulas 
for iron and steel members have been much modified from 
time to time. 

The earliest formula, on which those at present in use 
are founded, was Gordon's, which was derived from Hodg- 
kinson's experiments. The symbol /i denotes the least dimen- 
sion of the cross-section, or the dimension measured in the 
direction in which lateral flexure is most probable. This 
formula for iron columns of rectangular section was 

p = 36,000 s ^ ( I H ), 

the last term of which was changed by Rankine into the gen- 
eral form P -^ 36,ooor\ as /i^ = I2r^ for a rectangle. For 
other forms of cross-section the ratio of the least dimension 
/i to r would change. The two values 36,000 must not be 
confused; the numerical identity is accidental. 

Sometimes the ratio I ^ h was represented by H or some 
similar symbol, and the formula was P = / S -^ ( i + ^ H^) 



P = 8,oooS -^ fi H ,) 

V 4.0,000/" J 



SAFE WORKING STRESSES. I77 

where / varied from 42,500 to 36,000, and a from i -^ 5,820 
to I -^ 2,700 for columns with fixed ends and different 
forms of cross-section. 

These values were derived from columns tested to failure, 
and the allowable mean unit stress in compression was then 
obtained by dividing/ by 4 + 0.05H, thus making the allowed 
unit stress smaller as the column became more slender. 

Cooper's specifications for 1884 gave 

P 
40,000; 

for square ends, 40,000 being changed to 30,000 and 20,000 
for one and two pin ends respectively. No piember was to 
have an unsupported length of more than 45 times its least 
width. 

Jos. AI. Wilson in 1885 gave the permissible mean unit stress 

for wrought iron members in compression, f -^ { 1 -\- — ,^ \ 

V 36,ooo;'-y 

for both ends fixed, with a substitution of 24,000 and 18,000 for 
36,000, for the cases of one and two pin ends. But/ was a vari- 
able determined as follows: — For pieces subjected to compression 

■ -11 • ,- r , i^iii- stress'A , 

onlv, permissible unit stress / = bjcrool i 4- I for 

^ ^ -^ 'J V 1 ^^^^^ stressy 

rolled iron, and for pieces subjected alternately to tension and com- 

/ max. stress of lesser kind ~\ 

pression f =r 6,500 I i — --; — ^ I. The 

V. 2 . max. stress or greater kmdy 

required section for tension was to be found by either of the last 

two formulas, after substituting 7,500 for double rolled bars, and 

7,000 for plates and shapes, in place of 6,500. 

C. C. Schneider, in 1886, in a design for the arched bridge 

over the Harlem river, New York, proposed to use for compression. 

r f p-\ 

members, f ^ j i + -r^ • — , I, where 



'' -^ ^ 1/ + 8E • ?J' 

/- , min. stress \ 

/ .— 10,000 j I -f- I 

\ 2 max. stress^ 

for wrought iron members subjected to one kind of stress only, and 

/' min. stress "\ 

/ = 10,000 I I — I 

V 2 max. stressy 

for similar members subjected to alternate tension and compress- 
ion, no regard being paid to the sign of the stress. In the case of 
steel use 12,000 in place of 10,000. For tension members use the 
last two formulas alone for/. 
13 



178 STRUCTURAL MECHANICS. 

The Pencoyd Iron Works, in 1887, gave for wrought iron, 

14,000 -:- ( I + 5 I and 18,000 ^ ( i 4- -, I 

V i5,ooo;'V V^ io,ooo;'-y 

for steel, limiting a compression member to a length not exceeding 
120 times its least radius of gyration. The above formulas un- 
modified would apply to stresses from steady load. The rolling 
load stresses were to be increased by adding a variable percentage 
obtained as follows: — The effects of impact and vibration shall be 
considered and added to the maximum stresses resulting from 
engine and train loads. The effect of impact is to be found by 

multiplying the calculated max. live load stress by (0.7 + - ), 

i-d 

where L = number of panels loaded multiplied by a panel length 

(or loaded distance to produce the live load stress in question). 

Colorado Midland Railway Specifications. 

Main posts, chords and struts. 

-r^, J / ^ xT , ^^ii"^- stress~\ 

plat ends (7j5oo — 25 — ) i -f- I 

r V max. stress^ 

One pin end " 30 '' '' " 

Two pin ends '' 35 " " " 

"Lateral and other struts, subject to maximum stresses as cal- 

/ 
■culated, 11,000 — 45 — . 

J. A. L. Waddell, in 1887, for highway bridges under a live 
load of 100 lbs. per sq. ft., specified for truss members, 

12,000 — 4 c; — 

yfor iron. ^^ !>- for steel. 

One pm " 35 " ( '' 53 " \ 

Two pins " 40 '' J '' 60 '' J 

Cooper's specifications for 1890 — Railway bridges. 

Wrought iron. Medium steel. 

i I I . 

vChord ) S,ooo — 30 —^ 10,000 — 36 ~ for live load stresses. 

( ir),ooo — 60 " 19,000 — 72 " " dead " " 

( 7,000 — 40 " 8,500— 55 " " live " 

"I I4,')00 — 80 " 17,000 — no" "dead " •' 
( 10,500 — 5o " 13,000 — 85 " " wind " 

/ . 
Lateral struts 9,000 — 50 -7" f^^" assumed initial stresses. 

All compression pieces are restricted to an unsupported 
length of less than 45 times the least width. No distinction 
is made between pin and fixed ends. The mean unit stress is 
seen to be reduced for pieces whose loads are more rapidly or 
frequently imposed. 



Flat ends 9,000 — 30 



r 



segments. 

All 
posts 



SAFE WORKING STRESSES. 



179 



175. WorkingStresses: Compression. — Cooper's speci- 
fications for 1896 omit all requirements for wrought iron, and 
require for medium steel — 

( / 

^, , „ 1 10,000 — 45 — for live load stresses. 

Chord Segments -< tj ^ 

/ 20,000 — 90 " " dead 

Posts of J 8,500 — 45— "live 

Through Bridges') ,, ,, •, ■, 

^ ^ r 17,000 — go " " dead 

Posts of Deck I 9,000-40— "live 

Bridges and -I ^ ^ r 

Trestles | 18,000 — 80 " " dead 

( ^ 

Lateral Struts and J 13.000 — 70 — 

Rigid Bracing ) xr v 1 j ^ ^u- ^ c ' 
^ == / Jbor live load, two- thirds or same. 

Soft steel may be used in compression, with unit stresses 
I 5 per cent, less than those allowed for medium steel. 
A recent specification for railway bridges gives: 

Wrought Iron. Soft Steel. Medium Steel. 

/ / / 

Flat ends 8,600 — 27^- 10,000 — 33 ^~ 11,000 — 38 — 7- 

One pin " 30 " " 37 " " 43 " 

Two pins " 33 " " 4[ " " 48 " 

For highway bridges, increase 25 per cent. 
For roof trusses and buildings the mean unit compressive 
stress may be taken as. for 

Wrought Iron. 

/ 
Flat ends 10,750 — 33 — 

which values are the same as indicated above for highway 
bridges. 

For cast-iron, / may be taken as 10,000, and a as 
I _j_ 60,000. The use of cast-iron for compression members 
and beams is not approved. 

The compression flange of built beams and girders is 
usually made of the same gross section as that of the tension 
flange. 

176. Alternating Stresses. — Good practice requires a 
larger section to be given to members which are alternately 
subjected to tension and compression than would be needed if 
the stress were always of one sign. The discussion in § § i 56-8 
makes the reason plain. 



Soft Steel. 


Medium Steel. 


/ 
2,500 — 42 — 


/ 
13.750 — 48 ^. , 



l8o STRUCTURAL MECHANICS. 

Here, as well as in the specifications for columns, there 

has been much change, and the present tendency is to adopt 

the Launhardt-Weyrauch formula or an analogous one. 

An old specification for wrought iron members was — Required 
section = max. tension -^ 10,000 -|- max. compression -^ ^ col- 
umn strength, in case this result is larger than that given by the 
usual formula for columns. The phrase one-fourth column strength 
was intended to mean one-fourth of the unit stress given by the 
column formula when /"was taken as 36,00c to 40,000, § 174. 

The Pencoyd specifications say — Members subjected to 
alternate stresses of tension and compression shall be so pro- 
portioned that the total sectional area is equal to the sum of 
the areas required for "each stress. 

Theodore Cooper specifies — Members subject to alternate 
stresses of tension and compression shall be proportioned to 
resist each kind of stress. Both of the stresses shall, how- 
ever, be considered as increased by an amount equal to o. 8 of 
the least of the two stresses, for determining the sectional 
areas. 

Another engineer says — For compressive stress alone, 
use formula for posts. For the greater kind of stress, use 
max. lesser stress 
2 max. greater stress 

J. A. L, Waddell prescribes — In any portion of a bridge 

in which the stresses of tension and compression alternate, the 

sectional area required is to be determined by dealing with the 

part thus affected, — first for the calculated maximum tension, 

then for the calculated maximum compression, — employing 

for unit stresses the values that would be used were there no 

smaller stress a 

reversion of stress x ( i — h "i 1 I, and adopting the 

^ larger stress / r o 

greater of the two areas thus found. 

A recent specification for railroad bridges gives, for the 

greater stress, a value 

Wrought Iron. Soft Steel. Medium Steel. 

, max. lesser stress "\ • o ^: / j \ / 1 \ 

7,500 (i — I 8,000 (i — ao.) 9,400 (i — do.) 

2 max. greater stressy 

and for highway bridges 

9,400 (I — do.) 10,800 (i — do.) 11,700 (i — do.'' 

or 25 per cent. more. 



max. lesser stress ^ 

7,000 (I 

^ 2 max. greater stress J 



SAFE WORKING STRESSES. 



i8r 



In specifications for the Hudson River suspension bridge, 
Mr. Cooper gives, for the stiffening trusses — Under the rever- 
sal of stress by live loads, the chords of the stiffening trusses 
shall not be subjected to a greater unit stress per square inch 
than 

T 



unit tension = 



unit compression 



T + ^7 C 
a C 



20,000 



T + ^ C 



( 



20.000 



')-■ 



nor the web diagonals of the stiffening trusses to a greater unit 



stress than 



unit tension 



T 



unit compression 



I 4- ^C 
a C 



- 18,000, 



( 17,500 — 75 -j ; 



T + ^ c 

where the quantit}' in parenthesis is the column formula; 
T = total tension in the member; C = total compression in 
the member; and a — value of the net section in terms of the 
gross section of the member. 

177. Shearing and Bearing Stresses. — Working unit 
stresses in pounds per square inch. 



r-, ( Railway bridsfes 

bhear on 1 tt- u ' 1 • ? 

J . ^ A Hiffhwav bridges 

pins and rivets J -o r " j i -Tj- 

^ ( Roots and buildings 



Shear on 
webs of girders 

Bearing on 
diameter of pins 
and rivet holes 

Bending stress 
on pins. 



Railway bridges 
Highway bridges 
Roofs and buildings 

Railway bridges 
Highway bridges 
Roofs and buildings 

Railway bridges 
Highway bridges 
Roofs and buildings 



Wrought 
Iron. 


Soft Steel. 


^Medium 
Steel. 


7,000 


8,000 


8,700 


8,000 


9,000 


10,000 


9,000 


10,000 


11,000 


5.000 


5,700 


6,800 


5.500 


6,500 


7,500 


6,000 


7,000 


7.500 


12,000 


14,000 


15,000 


15,000 


17,000 


19,000 


18,000 


20,000 


22,000 


15,000 


17,000 


19,000 


17,500 


20,000 


22,000 


20,000 


22,500 


25,000 


12,000 


14.000 


16.000 



Bending stress on Purlins 

The number of field rivets should be 25 per cent, in 
excess ot that required for power riveting. 

For lateral connections 25 per cent, greater stress maybe 
allowed. 

For e3'ebar heads the stress in bearing may be 25 per cent, 
greater than that in the bar. 



l82 STRUCTURAL MECHANICS. 

178. Pedestals. — For bearing plates and pedestals, the 
working compression on blocks per square foot, in pounds, may 
be, for granite, 95,000; limestone, 80,000; sandstone, 45,000; 
brickwork, 6,700; concrete, 7,000. 

For rollers, not less than 2 in. in diameter, the pressure 
in pounds per linear inch of roller shall not exceed yooy' d for 
wrought iron, or goo^dior steel, <:/ being taken in inches. 

The pressures transmitted to the masonry by the pedestals 
shall not exceed 40,000 pounds per square foot under maxi- 
mum possible loading; and pressure within the masonry or 
upon rock foundation shall not exceed 20,000 pounds per 
square foot. 

The student is advised to study carefully two or more 
standard specifications for bridge or structural work, and to 
note the requirements for members and their details. The 
discussions in this book embody an attempt to make clear the 
reasons for such requirements. 



CHAPTER XI. 

INTERNAL STRESS: CHANGE OF FORM. 

179. Introduction. — Let any body to which forces are 
applied be cut by a plane of section. Stresses of tension, 
compression or shear, — normal, oblique or tangential,— may 
exist between the particles at the section. It is desirable to 
know the magnitude and kind of the unit stress at each point 
in order to be sure that the material can safely resist it; or to 
determine the required cross-section to reduce the existing 
stresses to safe values. 

A unit stress is expressed as a certain number of pounds 
of tension, compression, or shear on a square inch of section. 
If the plane of section is changed in direction, the force on 
the section may be changed and the area of section may also 
be changed, so that the unit stress on the new section is 
altered from that on the old in two ways. Stresses per square 
inch, or unit stresses, therefore cannot be resolved and com- 
pounded as can forces, unless they happen to act on the same 
plane. Generally, each unit stress may be multiplied by the 
area over which it acts, and the several forces so obtained may 
be compounded or resolved as desired; the final force or forces 
divided by the areas on which they act will give the desired 
unit stresses. 

Where the stress on a plane varies from point to point, 
as does the direct stress on the right section of the beam, and 
as does the shearing stress also in the same case, the investi- 
gation is supposed to be confined to so small a portion of the 
body that the stress over any plane may be considered to be 
sensibly constant. 

180. Stress on a Section Oblique to a Given Force. — 
Suppose a short column or bar to carry a force of direct 
compression or tension, of magnitude P, centrally applied and 
uniformly distributed over the cross-section S. The unit stress 
on and perpendicular to the right section will be /i = P -^ S. 



i84 



STRUCTURAL MECHANICS. 



On an oblique section C D, Fig. 2, making an angle o with 
the right section A B, the unit stress will be P -^ S sec a = 
/>! cos d, making an angle of O with the normal to the oblique 
section on which it acts. If this oblique unit stress is resolved 
normally and tangentially to C D, the 

Normal unit stress = /n = /i cos . cos ^ p^ cos^ d\ 
Tangential do. = ^ ^= /i cos sin 0. 

The normal unit stress on the oblique plane is of the 
same kind as P, tension or compression; the tangential unit 
stress, or shear, tends to make one part of the prism slide 
down and the other part slide up the plane. 

The largest normal unit stress for different planes is found 
when — o^ which defines the fracturing plane for tension; 

the minimum normal unit stress oc- 
curs for — 90°; and the greatest 
unit shear is found for — 45°, 
when we have q max. = \ p^. 
"d 181, Combined Stresses. — The 
^J___ action line of P maybe taken for the 
axis of X. Two equal and opposite 
forces, pull or thrust, may then be 
applied along the axis of Y, and the 
normal and tangential unit stresses 
found on the plane just discussed; 
and similarly for the direction Z. The normal unit stresses, 
since they act on the same area, may then be added algebra- 
ically, and the shearing stresses may be combined; finally a 
resultant oblique unit stress may be found on the given plane. 
A more convenient method will, however, be developed 
and used in the following sections. As most of the forces 
which act on engineering structures lie in one plane or parallel 
planes, such cases chiefly will be considered. 

182. Unit Shears on Planes at Right Angles.— If, in 
the preceding illustration, the unit stresses, both normal and 
tangential, are found on another plane N which makes an 
angle 0' = 90° — with the right section, there will result 

p'^ — p^ cos^ 0' — /, sin- 0\ q' — q. 




B 



.N 



}S-2. 



INTERNAL STRESSES. 



185 




Hence, on a pair of planes of section at right angles to 
one another the unit shears aj^e of equal magnitude, and, 
since p^ -f- p\ = p^ the unit normal stresses are together 
equal to the original normal unit stress. It is further evident 
that one normal unit stress p'^^ may be found by subtraction as 
soon as the other is known, and that ordinary resolution on 
these two planes of the original unit stress would be erroneous. 

183. Unit Shears on Planes at Right Angles Always 
Equal. — Since, as before stated, other forces, in other direc- 
tions, may be simultaneously applied to 
the given body, and their effects found 
on the same two planes, it follows /°2 
that, in any body under stress, the 2init 
shearing stress, on each of any two 
planes at right angles, will be equal : — 
a very valuable principle. 

Example.— \. closed cylindrical re- Fi^. 66. 
ceiver, J- in. thick, has a spiral riveted 
joint making an angle of 30° with the axis of the cylinder, and a 
portion 2 in. X 4 in. of the cylinder, Fig. 66, has the given tensions 
of 2,500 lbs. acting upon it. Then /j = 2,500 -4- 2 . ^ = 5,000 
lbs. per sq. in., and/o = 2,500 -^ 4 , ^ — 2,500 lbs. per sq. in. 

A = S'Ooo • J + 2,500 . J = 4,375 ll^s. per sq. in. 

,q z= 5,000 . 0.433 — " 2,500 . 0.433 ^ 1,082 lbs. per sq. in. 

/ = 1/(4,375' + 1,082') ^ 4,507 lbs. per sq. inch., 

or 4,507 . ^ = 1,127 lbs. per linear inch of joint, which value will 
determine the necessary pitch of the rivets for strength. 

The stress on a joint at right angles to the above can be simi- 
larly found. An easier process will be given in § 190. 

184. Compound Stress is the internal state of stress in 
a body caused by the combined action of two or more simple 
stresses (or balanced sets of external forces) in different direc- 
ti<^ns, as in the above example. The investigations which 
follow are those of compound stress, but they will, as above 
stated, be chiefly confined to stresses in or parallel to one 
plane. 

185. Conjugate Stresses: Principal Stresses. — If the 
stress on a given plane in a body is in a given direction, the 
stress on any plane parallel to that direction must be parallel 
to the first-mentioned plane. For the equal resultant forces 



i86 



STRUCTURAL MECHANICS. 




Ficf.eJ. 



exerted by the other parts of the body on the faces A B and 
C D of the prismatic particle, Fig. 6j , are directly opposed to 
one another, their common hne of action traversing the axis 
of X through O; and they are therefore independently bal- 
anced. Therefore the forces exerted by the other parts of the 
body on the faces A D and B C of this prism 
must be independently balanced and have their 



y resultants directly opposed; which cannot be 
unless their direction is parallel to the plane 
Y O Y. 

A pair of stresses, each acting on a plane 
parallel to the direction of the other, is said 
to be conjugate. Their unit values are inde- 
pendent of each other, and they may be of the same or 
opposite kinds. If they are normal to their planes, and hence 
at right angles to each other, they are called principal stresses. 

Exa7nples. — The unit stress found in § 183 makes an angle 
with the plane on which it acts whose tangent is 4,375 -r- 1,082 = 
4.04. Upon a new plane cutting the metal in this direction the 
stress must act in a direction parallel to the joint referred to. 

If a plane be conceived parallel to a side-hill surface, at a 
given vertical distance below the same, the pressure at all points 
of that plane, being due to the weight of the prism of earth above 
any square foot of the plane, will be vertical and uniform. Then 
must the pressure on a vertical plane transverse to the slope be 
parallel to the surface of the ground. That the pressure against 
the vertical plane is not horizontal, but inclined in the direction 
stated, is shown by the movement of 
sewer trench sheeting and braces, when 
the braces are not inclined up hill, 
but are put in horizontally. 

186. Shearing Stress.— If the 

stresses on a pair of planes are 
entirely tangential to those planes, o^ 
the unit shears must be equal. Con- 
sider them as acting along the faces of a small prismatic 
particle A B C D, which lies at O. The moment of the total 
shear on the two faces A B and C D must balance the moment 
for the faces A D and B C, for equilibrium. 

^'.AB.E F = ^.AD.HG. 
But the area of A B C D, A B . E F = A D . H G; . • . ^.' = ^ . 




"^^ .Fig.68. 



INTERNAL STRESSES. 



187 



This construction shows further that a shear cannot act 
alone as a simple stress, but must be combined with an equal 
unit shear on a different plane. 

187. Two Equal and Like Principal Stresses. — If a 
pair of principal stresses, § 185, are equal unit stresses of the 
same kind, p^ and /o , Fig. 69, the stress on every plane is 
normal to that plane, and of the same kind and magnitude. 

Let /i act in the direction O X on the plane O' B' of the 
prismatic particle O' A' B' which lies at O. and /a act in the 
direction O Y on the plane O' A', p^ and p^ being equal unit 
stresses of the same kind. Make O D = /i . O' B', the total 
force on O' B', and O E = /a . O' A', the total force on O' 
A', both being positive. Complete the rectangle O D R E. 
Then must R O, applied to the plane A' B', be necessary to 
insure equilibrium of the prism O' A' B'. Hence /' = R O 
-^ A' B'. Since ^j = ^2 » 



O D 



O E 



O R 



O' B' O' A' 



A' B 



t ^ 



f = Px= A 



Because of similarity of triangles A' O' B' and O E R, R O is 




\v\ \ ri^.70. U 



B Fig,. 69. 



perpendicular to A B, or p' to A' B', and is of the same kind 
as /i and p^. 

Example. — Fluid pressure is normal to every plane passing 
through a given point, and equal to the pressure per square inch 
on the horizontal plane traversing the point. Here manifestly the 
three co-ordinate axes of X, Y and Z might be taken in any posi- 
tion, as all stresses are principal stresses. 

188. Two Equal and Unlike Principal Stresses. — 
If a pair of principal stresses are equal unit stresses of oppo- 
site kinds, as p^ and — p^, Fig. 70, the unit stress on every 
plane will be the same in magnitude, but the angle which its 
direction makes with the normal to its plane will be bisected 



I 88 STRUCTURAL MECHANICS. 

by the axis of principal stress, and its kind will a^ree with 
that of the principal stress to which it lies the nearer. 

In this case lay off Oe in the negative direction, to repre- 
sent ■ — ■/., . O" A"; construct the rectangle O D/r, and draw rO 
which will be the required force distributed over A" B" to 
balance the forces O" B" and O" A". This force rO will be 
the same in magnitude as R O, making p' = p^ = p,^ and rO 
will make the same angle with O X or O Y as R O does. As 
R O lies on the normal to A B, and O X bisects R Or, the 
statement as to position is proved. The stress p agrees in 
kind with that one of the principal stresses to which its direc- 
tion is nearer. 

189. Two Shears at Right Angles Equivalent to an 
Equal Pull and Thrust.— If the plane A" B" is at an angle 
of 45° with O X, rO will coincide with A B and becomes a 
shear. Therefore two equal unit stresses of opposite kinds, 
that is a pull and a thrust, and normal to planes at right 
angles to one another, are equivalent to, and give rise to, 
equal unit shears on planes making 45° with the first planes 
and hence at right angles to 
each other; and vice versa. 



%\ 



Example. — If, at a point in Yj 

the web of a plate girder, Fig. 71, 



Ij 



1^ 




there is a unit shear, and nothing p-j - j* 

but shear, on a vertical plane, 

of 4,000 lbs., there must be a unit shear of 4,000 lbs., and noth- 
ing but shear, on the horizontal plane at that point; and on the 
two planes inclined at 45° to the vertical through the same point 
there will be, on one, only a normal unit tension of 4,000 lbs., 
and on the other an equal normal unit thrust. 

From a combination of the two preceding demonstra- 
tions follows the more general problem. 

190. Stress on any Plane, when the Principal Stresses 
are Given. — Let the two principal unit stresses be /j = O D, 
and p-i = O F, of any magnitude, and of the same kind or 
opposite kinds. Fig. 72. The direction, magnitude and kind 
of the unit stress on any plane A B is desired. 

Let /i be the greater. Divide p^ and p., into their half 
sum and difference as follows: — 

A = M/i +A) + MA— A). andA = ^(A4-A)— i(A— A)- 



INTERNAL STRESSES. 



189 



The distance O C or O E will represent the half sum 
\{P\ + AO' 3-iTd C D or E F the half difference J( P^ — //). 
If /i and /o are of the same si^n the right hand figure will 
result; if of opposite signs, the left hand figure will be 
obtained. 

R}' the principle of § 187, when the two equal principal 
unit stresses O C and O E are considered, lay off O M on the 
normal to the plane 
whose trace is A B, 
for the direction 
and magnitude of 
the unit stress on 
A B due to 

i(A +A0. 

There remain C D 
and E F represent- 
ing + i(A —A) 
on the vertical axis, 
and — \{ p, — /,) 
on the horizontal 
axis respectively. 

By § 188, lay off O Q, making the same angle with O X 
as does O M, but on the opposite side, or in the contrary 
direction, for the magnitude and direction of stress on plane 
A B due to + J(/i — /o). As O M and O Q both act on the 
same unit of area of A B, R O, in the opposite direction to 
their resultant O R, will give the direction and magnitude of 
the unic stress on A' B' to balance /i on O' B' and/o on O' A'. 
In the figure on the right R O is positive, or compression. If, 
in the figure on the left, where /] is + and /,, is — , R O falls 
so far to the right as to come on the other side of A B, it will 
agree with/.,, and be negative or tension. If A B is taken 
much more steeply inclined, such will be the case. The small 
prisms illustrate the constructions. If R O falls on A B, it 
will be shear. Some constructions for different inclinations 
of plane A B will be helpful to an understanding of the matter. 

A much abbreviated construction is as follows: — Strike a 
semicircle from M on the normal, with a radius M O = ^(/j +/.,), 
and draw M R through the points where the semicircle cuts 




190 STRUCTURAL MECHANICS. 

the axes of p^ and p.i. The ano^le N M R is thus double the angle 
MOD, since it is an exterior an^le at the vertex of an isosceles 
triangle. Lay off M R =■ \{Px — A) in the direction of the 
axis of greatest stress, and R O will be the desired unit stress 
on A B. If A is opposite in kind to /i, M R will be greater 
than M O, and R will go beyond P. 

igi. Ellipse of Stress.— For different planes A B 
through O, px and A being given, the locus of M is a circle of 
radius J(/i -\- pi), and the locus of R is an ellipse (as will be 
proved below), with major and minor semidiameters /i andA- 
Hence it is seen why /i and A* normal to the respective planes, 
and at right angles, are called principal stresses. 

If three principal stresses, coinciding in direction with the 
rectangular axes of X, Y and Z, simultaneously act on a given 
point, an ellipsoid constructed on them as semidiameters will 
limit and determine all possible stresses on the various planes 
which can be passed through that point in the body. 

That the locus of R is an ellipse may be proved as fol- 
lows: —Drop a perpendicular R S from R on O X. P M O 
and O M G are isosceles triangles. <POM = <GOB = ^. 

OM = MP = i(A +A); GR =.A; PR = /,. 

R S = P R sin M P O = A sin P O M = A sin 0, 

S O =^ G R sin M G O == A cos P O M =: A cos ^, 

O R = A = l/(S O'^ + R S^) = i/(A' cos' + A, sin'^ 0). (i.) 

which is the value of the radius vector of an ellipse, the origin 
being at the centre. 

AS<NMR = 2.<P0M=:2^, 

sin N O R : sin O M R = R M : O R = |( p, — j)^) . ^r 



'r ; 



.-. sin N O R = sin 2 ^ . -^^^ ^, (2.) 

which gives the obliquity of the unit stress to the normal to 
the plane, in terms of the angle of the normal with the axis 
of greater principal stress, or of the plane with the other axis. 
The graphical construction gives the stress and its angle with 
the normal or the plane by direct measurement, and is far 
more convenient and less liable to error. 

\{ p^ = p.^, the case reduces to that of § 187 or 188. 



INTERNAL STRESSES. I9I 

If the ellipse whose principal semi-diameters are/i and/2 
is given, the unit stress on any plane may briefly be found by 
drawing the normal \,o the plane, laying off O M = ^ (/i + p^, 
taking a radius of \ {p^ — /s)- and, with M as a centre, cutting 
the ellipse at R on the side of the normal towards the greater 
axis. A line R O will be the desired unit stress. 

Exainple. — Let/^ = 100 lbs. on sq. in.,/2 = — 5° lbs. Plane 
A B makes 30° with direction of p^, or its normal makes 30° with 
^,. Construct the figure and find the magnitude, direction and 
sign of the unit stress on the oblique plane. Try other values. 

192. Shearing Planes. — To determine the angle of the 
planes on which there is only shear, and the conditions which 
render shearing planes possible. 

If the plane A B of the previous figure is to be the shear- 
ing plane, there must be no normal component upon it, and 
therefore, from § 180, if the plane makes an angle 0^ with 
p2. or the normal to it is inclined at an angle Q^ to p,, 

/,! = p^ cos^ ^s + A sin^ 6*3 = o. 

sin 6's r — p\ 

. • . ^ = tan (9s = V — ^ . 

cos^s ^V p^J 

No shearing plane is possible unless j)^ and p, differ in 
sign. There will then be two planes of chear making equal 
angles with the direction of ^2 or of pi. 

In the above example, \/(ioo -^- 50) = ^2 =. tan 6^ = 
1. 4142. d =r. 54° 44'. 

If the ellipse of stress is drawn, take a radius equal to the 
side of a right angled triangle whose other side is h,{pi + jo-zjj 
and hypothenuse is J(_P] — P2), and strike a circle from the 
centre of the ellipse. Planes drawn through the points of 
intersection of this circle and ellipse and the centre will be the 
shearing planes. Unless p, and P2 differ in sign, the circle 
will be imaginary. The value of the shear on these planes is 

^ = V(i(/. -Af - KA + A)0 = VaA. 

193. Given any two Stresses: to find Principal 
Stresses.- As, in actual practice, two oblique unit stresses on 
different planes may often be known in magnitude, direction 
and sign, it will then be required to find the principal unit 



192 STRUCrURAL MECHANICS. 

stresses, since one of them is the maximum stress to be found 
on any plane, and the other is the minimum stress of the same 
kind, or the maximum normal stress of the opposite kind. 

Given two existing unit stresses, p and p', of any direc- 
tion, magnitude and sign, to determine the principal unit 
stresses, p^ and p^. 

If pi and P2 were known, and p and p' were th n to be 
found from the former, the construction shown in Fig. 73 
would be made, in which O M = O M' = h{P\ + p-i) ^.nd 
M R = M' R' = ^{px — p-?). If one of these normals were 
revolved around O to coincide with the other, the point M' 
would fall on M, but M' R' would diverge from M R, while 
equal in length to it. 

Hence, when p and p are the given quantities, let A B, 
Fig. 74, represent the trace of the plane on which p acts, O N 
the normal to that plane, and O R the unit stress p in magni- 
tude and actual direction of action on A B. OR represents 
either tension or compression, as the case may be. Now let 
the plane on which p' acts, together with its normal and 
p' itself in its relative position, be revolved about O until it 
coincides with A B. Its normal will fall on O N and p)' will 
be found at O R', on one side or the other of O N, if it is of 
the same kind with p; or it is to be laid off on one of the dot- 
ted lines below, if of the opposite sign. 

In other words, lay off p' from O, at the same angle with 
O N which it makes with the normal to its own plane. It is 
well, for accuracy of construction, to draw it on the same side 
of the normal as p, the result being the same as if it were 
drawn on the other side. (The change from one side of the 
normal to the other simply consists in using the corresponding 
line on the other side of the main axis of the ellipse of stress). 
Thus is found O R' or — O R' as the case may be. Draw 
R R' and, since R M R' must be an isosceles triangle, bisect 
R R' at T and drop a perpendicular to R R' from T on to the 
normal, cutting the latter at M. Then since, as previously 
pointed out, O M = ^{p, -f p.,) and M R = M R' = 
i (pi — /)2), — with M as a centre, and radius M R, describe a 
semicircle; O N will be pi and O S will be p.,. Since p is in its 
true position, and the angle N M R = 2 M O D of Fig. 72 or 



INTERNAL STRESSES. 



193 




»'i&74. 



2 M O X of Fig. 73, the direction of the axis X along which 
/i acts will bisect N M R, and the axis along which p^ acts 
will be perpendicular to axis X. They may be now at once 
drawn through O, if desired. 

194. From any two Stresses to find other Stresses. — 
From the preceding construction, § 193, the stress on any 
other plane may now 
be found. All pos- 
sible values of p con- 
sistent with the two, 
O R and O R', first 
given, will terminate, J^^ 
in Fig. 74, on the '^ ^ 
semi-circle just ^ ^t7^ "^ o^B 
drawn, as at R", and the greatest possible obliquity to the 
normal to any plane through O will be found by drawing 
from O a tangent to this semi-circle. 

195. When Shearing Planes are Possible. — In case 
the lower end of the semi-circle cuts below O, Fig. 75, p^ and 
p2 are of opposite signs, all obliquities of stress are possible, 
a"-nd the distance from O to the point where the semi-circle 
cuts A B, being perpendicular to the normal O N, gives the 
unit shear on the shearing planes. If p^ and p^ are drawn 
through O in position, and the ellipse of stress is then con- 
structed on them as semi-diameters, (as can be readily done 
by drawing two concentric circles with p^ and /g respectively 

as radii, and projecting at right angles, 
parallel to pi and /s, to an intersection, the 
two points where any radius cuts both 
circles), an arc described from O, with a 
radius equal to this unit shear and cutting 
the ellipse, will locate a point in the shear- 
ing plane which may then be drawn through that point 
and O. Two shearing planes are thus given, as was proved 
to be necessary, § 186. 

The above solution may be considered the general case. 

196. From Conjugate Stresses to find Principal 
Stresses. — If / and/' are conjugate stresses, it is evident, 

from definition, and from Fig. Sy,' that they are equally 

14 




194 



STRUCTURAL MECHANICS. 



inclined to their respective normals. Hence O R' will fall on 
O R, when revolved, both O R and O R' lying above O when 
of the same sign, and on opposite sides of A B when of oppo- 
site signs. The rest of the construction follows as before, 
being somewhat simplified. When one conjugate stress is 
shear, the other is shear, of the same unit value, by § i86, 
but the stress on any other plane cannot be shear. 

It may be noted that, when /^ = + /,, the propositions of 
§ § 187 and 188 are again illustrated. 

One who is interested in a mathematical discussion of this 
subject is referred to Rankine's '^\pplied Mechanics," where 
it is treated at considerable length. This graphical discussion is 
much simpler, less liable to error, and determines the stresses in 
their true places. 

For the application to the pressure of earth against a wall or 
any plane see § 255. 

197. Change of Principal Stresses by a Shear. — The 

varying tensile and compressive stresses on any section of a 



r- ^ 

iD' ^-- 




Pyt 



^■•'^76. 



M 



beam are accompanied by varying shearing stresses on that 
section and by equal shears on a longitudinal section. The 
direct or normal stresses due to bending moment vary uni- 
formly from the neutral axis either way, § 75; the shears are 
most intense at the neutral axis, § 86. The maximum and 
minimum stresses on any particle with their directions may be 
found as follows: — 

Given the normal and tangential components of the 
stresses on two planes at right angles, to find the magnitude 
and direction of the new principal stresses. This problem is 
a special case of § 193, ^ and/' being given by their com- 
ponents. Let the original normal unit stress, due to the 
bending moment, and normal to the section A B, Fig. yG, be 
denoted by/^ • I^ there is any normal stress on the plane at 
right angles to A B, denote it by/y'. See § § 198, 206. The 
two equal unit shears on the two faces will be ^. 

Lay off /x = O N on the normal to the plane A B om 
which it acts in its true direction. As q acts on the same area 



INTERNAL STRESSES. 



195 



of A B as does p^ , lay off N C = ^ at right angles. The two 
component stresses being O N and N C, a line from O to C 
will represent the direction and magnitude of the unit stress 
on A B. Revolve the plane on which /y acts through 90°, to 
coincide with A B, so that the normals coincide. Then will 
/y fall at O F, if of the same sign as p^ , or at O F' if of the 
opposite sign. F D or F' D' is q, the revolved shear, laid off 
in the direction opposite to that of N C, as its direction 
requires, — see sketch on the right. Then a line from O to D 
or D' will be the revolved stress. As O C and O D represent 
/ and/', by § 193, connect D with C; bisect D C at E, which 
point falls on O N and is also the point w,here the perpen- 
dicular from D C will strike O N. Hence 



o E = i(/,+/y) =.i(A +A); 

EC ^\^p,-p,) = ;/(EN^ + Ne) = T/(i(A 
Add and subtract. 



■A J + q\ 



/, :::= O E + E C =: |(^,+/y) + l/(i (A "A Y + /), 

'^^ ^ O E — E C = J (A + A ) - T/(i (A -A r + ^')- 

The new major principal unit stress p^ will bisect the 
angle NEC, and the new minor principal stress A will be at 
right angles to py Therefore 

Tan 2l!^ = tanNEC = ^-^l (A — A )' 

from which <?, the obliquity to O N, can be found. The two 
new principal stresses with their planes are represented in the 
figure. 

If E C is less than O E, A will have the same sign as A; 
if greater, A will be opposite in sign to p^. 

Example. — If /x = 5? 000 lbs., p^ = 1,000 lbs., ^ = 3,000 
lbs., then/i = 7,200, A — — 600, = 28° 09'. If/x = — SjOoo 
lbs., the values change to 2,440, — 6,440 and 22° 30'. 

Bending and torsion combined give rise to a construction 
for maximum stress like Fig. y6, A being usually taken equal 
to zero. There results /i =: i A + 1/(5 Px^ + ^^). from which 
value of max. stress is derived, in § 93, the max. necessary 
resisting moment or the required section of a beam or shaft 
subjected to bending and torsion together. 



196 STRUCTURAL MECHANICS. 

198. Local Loading. — The above investigation also ap- 
plies to the problem of local loading on a beam. In the pre- 
ceding Fig. 76, /y may denote the unit pressure on the hori- 
zontal plane at a certain point in a beam due to a heavy local 
load, if the load is on top, or a tension if the load is below the 
beam. If the load is applied elsewhere in the vertical, there 
will be pressure on horizontal planes below and tension on 
those above the load. 

As to the magnitude of p^ : — A load W at any point of 
the beam goes to the two points of supports as shear. This 
shear is distributed over the cross-section of a beam according 
to the law developed in § '^6. The pressure, or tension, on 
any horizontal plane F E, Fig. 35, p. 75, at a loaded point, 
will be that due to W less the amount of shear carried by the 
section of the fibres G E or E A on that side of this horizontal 
plane on which W is placed. It is doubtful whether the 
effect is of much importance so long as W does not injure the 
material. 

199. Axes of Direct Stress in a Beam. — The stress at 
one edge of any right section of a beam will be normal and 
compressive; at a point a little nearer the neutral axis, the 
normal stress will be a little less and a small unit shear will 
also be found, with the result that the direction of the real 
principal stress at that point is slightly inclined to the plane 
of the section. The direct stress decreases regularly as the 
centre is approached, the unit shear increasing, § 86, and at 
the neutral axis there is only shear on two planes at right 

angles, one longitudinal, the other 
transverse. These shears are equiva- 
lent to normal tension and com- 
'^'^^^- pression on planes at 45° with the 

axis. Hence the tangents at different points of curves such 
as represented in Fig. 77 will show the direction of the 
principal tension or compression at such points. 

At the section of max. bending moment, at which section 
the shear will be zero, the curves will be horizontal. They 
will cross the neutral axis at 45° and will be vertical when 
they reach the opposite side of the beam. The stress dimin- 
ishes along the curve, being maximum at the section of maxi- 




CHANGE OF FORM. 



197 



mum bending moment, equal to the unit shear at the neutral 
axis, and zero at the edge of the beam. 

200. Change of Volume. — If/ = unit stress per square 
inch on the cross-section of a prism, and X is the resulting 
stretch or shortening J?er unit of length, then by definition 
E = / -f- A, if / does not exceed the elastic limit. 

When a prism is extended or compressed by a simple 
longitudinal stress, it contracts or expands laterally, Fig. 78. 
This contraction or expansion per unit of breadth may be 
written ^ l -^ m, where i -^ m, the ratio of lateral contrac- 
tion to longitudinal extension, is a constant for a given 
material, and for most solids lies between 2 and 4. § 205. 

A simple longitudinal tension / then accompanies a 

Longitudinal stretch = A = / -^ E per unit of length, and a 
Transverse contraction = — X -^ m ■= — p -^ mY, per unit 
of breadth. 

For ordinary solids k is so small that it makes no difference 
whether it is measured per unit of original or per unit of 
stretched length. The original length will be used here. 

The new length of the prism is /(i + '^) and the cross- 
section is S(i — A -^ my. The volume has changed from S/ 
to S/( I -\- I — 2/ -:- ni) nearly, if higher powers of A than the 
first are dropped, since the unit deformations 
are very small. The change of unit volume is 

2 
therefore x ( i ). Thus, if m is nearly 4, 

for metals, the change of volume of one cubic 
unit is i I nearly, the volume being increased ^ ^u 
for longitudinal tension. If there were no 
change of volume, m would be 2, as is the case 
for india rubber, for small deformations. Similarly, for com- 
pression the change of unit volume is nearly — \X for metals, 
the volume being diminished. 

Example. — Steel, E = 29,000,000; 'p = 20,000 lbs. per sq. 
in, tension; the extension will be 




29,000,000; p 

of its initial length, the 



lateral contraction will be about 



i>45o 
I 

5, 800 



of its initial width, and its 



increase of volume about 



2,900 



198 



STRUCTURAL MECHANICS. 



201. Effect of two Principal Stresses.— Denote the 
stresses by pi and p2» treated as tensile. If they are compres- 
sive, reverse the signs. 

Under the action oi p^ there will be the following stretch 
of the sides per unit of length: — Fig. 79. 




Parallel to O C, §; 
E 



Pi 



parallel to O B and O A, ^. 

Under the action of pg there will be 



f^igTQ 



Parallel to O B, ^; 



parallel to O A and O C, - 
Adding the parallel changes or stretches 

Parallel to O C, /l^ = ^fp^—^^X, 
Parallel to O B, /I2 = ^ (b — - ); 



P2 



Parallel to O A, /I3 = 



mK 



{Pi 4- P2)' 



If 2h a-nd P2 are equal unit stresses, but of opposite signs, 
the changes of length become 






P I 

-^(i 4- ■ — ): and zero; 

E ^ 7n ^ 



or, putting either of these two changes equal to x, the lengths 
of the sides of the cube originally unity per edge will be i -\- X, 
I — X and I, and the volume, neglecting A^, is unchanged. 

202. Effect of two Shears. — By § 189, two equal prin- 
cipal stresses of opposite signs are equivalent to two unit 
shears of the same amount per square inch, on planes at 45° 
with the original axes. Hence the above distortion results 
from two shears at right angles, and necessarily equal; and 
such shears cause no change of volume. 

Fig. 80 shows the distorted cube. A square traced on 
the side of the original cube will become a rhombus, the angles 



CHANGE OF FORM. 



199 



of which are greater and less than a right angle by the equal 
amount 0. Now one-half the angle \~ — d has for its tangent 
i(i — ^) -i- 4(i + ^05 hence 



I 



I -\-k 



= tan 1(1- 



I — tan ^ 6 

I + tan 1 e' 



or A = tan -J 6. 



But as d is small, I :=z ^ d, ox d ^ 2X. 



Therefore a stretch and an equal shortening, along a pair 
of rectangular axes, are equivalent to a simple distortion rela- 
tively to a pair of axes making angles 
of 45° with the original axes; and 
the amount of the distortion is 
double that of either of the direct 
changes of length which compose 
it. This fact also appears from the 
consideration that a distortion of a 
square is equivalent to an elongation of one diagonal and a 
shortening of the other in equal proportions. 




For steel, as before, I = 
Pi = — ^2 = 20,000 lbs. 



1,450 



725 



= 4' 45", if 



203. Modulus of Shearing Elasticity. — Similarly, equal 
shearing stresses q on two pairs of faces of a cube, in direc- 
tions parallel to the third face, will distort that third face into 
a rhombus, each angle being altered an amount 6, there being 

Oi > distortion of shape only, and not change of 

; volume. Fig. 81. 

f Under the law which has been proved 

;' true within the elastic limit, and the definition 
/"^ of the modulus of elasticity, § 10, a modulus 
of transverse (or shearing) elasticity, C, also 
called co-efficient of rigidity, as E may be called co-efficient of 
stiffness, may be written, C = q -^ d. 

As these two unit shears are equivalent to a unit pull and 
thrust of the same magnitude per square inch, at right angles 




200 STRUCTURAL MECHANICS. 

with each other and at 45° with these shears, the case is iden- 
tical with the preceding one. Then 

e = 2X, and A =%(i H ). .-. 6 = ^. ^ . 

But, sisp = ^=Cd, C = — = ^ 



VI -j- I 



For iron and steel m is nearly 4, which gives C =: IE. For 
wrought iron and steel, C is one or two one-hundredths less than 
0.4E. Some use fE. C = 11,000,000 is a fair value. § 205. 

204. Stresses Resolvable into Shear and Fluid Press- 
ure. — All systems of stress acting on a body may be resolved 
into distorting or shearing stresses, which do not alter the vol- 
ume, and a stress 'p like fluid pressure, equal in all directions 
and normal, but positive or negative. 

Suppose a cube of unit length of side acted on by a nor- 
mal stress pi on two opposite faces. It will in no way alter 
the conditions of stress to apply + V ^^^^ — _p normally to each 
of the four remaining faces of the cube, and to make p = ipu 
as seen in Fig. 82. A + p on each horizontal face and a — p 

on two opposite vertical faces together form 
a pair of shearing or distorting stresses; 
another + p on each horizontal face and 
a — p on the other two vertical faces 
^ act similarly. These shears produce no 
change of volume. There remains, there- 
fore, a stress of + _p = Jp^, normal on 
rig.8£.^=N-3^ every face, in all constituting a fluid 

stress, which will increase or diminish the 
volume, according to the direction of pj. Other stresses on the 
other faces may be looked at in the same light. Since a body 
has three principal axes of stress, take the cube parallel to 
these axes. 

Example. — If p^ = 300, 'p<^ = — 180 and ^3, in the third co- 
ordinate direction = — 120 lbs. per sq. in., p = ^(300 — 180 — 
120) = o, and there will be no change of volume. If jOj = 450 
lbs. pressure, ^2 = i'a = 125 lbs. pressure, the change of volume 
per cubic inch will be due to 233J lbs. per sq. in. on each face. 



7>.^ 




CHANGE OF FORM. 



201 



205. Coefficient of Elasticity of Volume. — Let v be 

the change of volume per cubic inch of the body under a 
normal unit stress / on any and all areas. 

Then Y^ = p ~ v \?> called the coefficient of elasticity of 
volume. The relation between K and the other constants can 
now be easily found. 

A simple normal stress 3/, by § 200, will produce a 

change of volume 3A I i — — I. But, the shears being dis- 

carded, a simple normal stress p^ = 3/ produces the same 
change of volume as a normal stress / = \p^ on any and all 
areas of the cube. Hence 



2 f) 

K = ^ -^ 3/1(1 ); or, since A = - , 

m „ ni -\- \ 

K = E -. But, by § 203, E = 2C ; 

2,111 — 6 ;;/ 

^^ + 1 6K + 2C 9CK 

2,m — 6 3K — 2C 3K -|- ^ 

K C E m 

Bj-ass J 14,250,000 j 4,900,000 j 13,500,000 

/ 15,400,000 ] 5,700,000 I 20,000,000 ^' 

i- ' \ 6,300,000 \ 16,700,000 „ r 

Copper 23,000,000 J 'J ' J " ' 2.0 

^^ • :)'y J ^ 7,000,000 I 17,500,000 

Cast Iron 13,700,000 7,500,000 19,000,000 3.7 

Wrought Iron 20,700,000 11,000,000 28,000,000 3.6 

CI c c \ 29,000,000 „ ^, 

Steel 26,200,000 11,600,000 ] -4,800,000 ^^^ 

Timber \ 75,ooo f-^'sooiooo 
{ 100,000 ( 1,400,000 

C may be found from the torsional vibrations of a wire. If 
/ = time of a single oscillation in seconds, I =■ moment of inertia 
of the vibrating system about its axis of rotation, and T = twist- 
ing couple, t = "-j/(I d -^ T). Then §§ 89, 91, 

T = i-TT^i^,'; ^^ = Cr, -, T = l-Ci\' -^, and C = -^. 

206. Stress on One Plane the Cause of Other 

Stresses. — The elongation produced by a pull, the shortening 
produced by a thrust, and the distortion due to a shear.can be 
laid off as graphical quantities and discussed as were unit 
stresses themselves. All the deductions as to stresses have 
their counterparts in regard to changes of form. There has 



202 STRUCTURAL MECHANICS. 

been found an ellipse of stress for forces in one plane, when 
two stresses are given. Also, when three stresses not in one 
plane are given, there is an ellipsoid of stress which includes 
all possible unit stresses that can act on planes in different 
directions through any point in a body. So there is an ellipse 
or ellipsoid that governs change of form. 

Whether the movement of one particle towards, from or 
by its neighbor sets up a resisting thrust, pull or shear, or the 
application of a pressure, tension or shear is considered to 
cause a corresponding compression, extension or distortion, 
the stresses and the elastic change of form coexist. Hence 
it follows that, when a bar is extended under a pull and is 
diminished in lateral dimensions, a compressive stress acting 
at right angles to the pull must be aroused between the parti- 
cles, and measured per unit of area of longitudinal planes, 
together with shears on some inclined sections. 

That such a state of things can exist may be seen from 
the following suggestions. It may be conceived that the 
particles of a body are not in absolute con- 
tact, but are in a state of equilibrium from 
mutual actions on one another. They resist 
with increasing stress all attempts to make 
them approach or recede from each other, 
and, if the elastic limit has not been exceeded, 
they return to their normal positions when the 
Pig Q-z external forces cease to act. The particles 

in a body under no stress may then be con- 
ceived to be equidistant from each other. The smallest 
applied external force will probably cause change in their 
positions. 

If, in the bar to which tension is to be applied, a circle 
is drawn about any point, experiment and what has been 
stated about change of form in different directions show, that 
the diameter in the direction of the pull will be lengthened 
when the force is applied, the diameter at right angles will be 
shortened, and the circle will become an ellipse. In Fig. 83, 
particle i moves to i', 2 to 2', 4 to 4' and 7 to 7'. As they 
were all equidistant from o in the beginning, i in moving to 
i' offers a tensile resistance, 7 resists the tendency to approach 




CHANGE OF FORM. 203 

o, while a particle near 4, moving to 4', does not change its 
distance from o, but moves laterally, setting up a shearing 
stress. A sphere will similarly become an ellipsoid. 

207. Actual Resulting Stresses. — Let — p^ be the unit 
tension in the direction o-i, and + P2 the accompanying unit 
thrust in the direction 7-0. If a pull is applied to the solid in 
the direction 0-7, which develops — p\ tension in that direc- 
tion on the plane o-i, and p'j thrust in the direction i-o, the 
resultant unit tension along o-i on the plane 0-7 will be 
— Pi + p'l, and along 0-7 on the plane o-i will be — p'2 + P2' 
It follows that the tension in the direction o-i will be less 
than when the first pull was acting alone. 'Hence a plate is 
stronger to resist two pulls at right angles than when subjected 
to one only. The opposite deduction can be drawn if one 
principal stress is of opposite sign to the other. 

A boiler plate has a tension in a tangential direction, that 
is, on a linear inch of longitudinal element, oi pr, or of/r -^ t 
per square inch, where p = steam pressure per square inch, 
r = radius in inches, and t = thickness of plate. On a cir- 
cumferential inch the pull is one-half as much. Then, by 
§ 200, and Vv^hat has been stated above, 

— A= P^^ Pi — P^^ — ^^i = iPi = i pr, 

— P\ = ipr, P\ — \p'-2 = \pr. 

Hence pr — \ pr =1 | pr, or the true unit tension is less 
than the apparent tension by \2\ per cent., and the boiler is 
stronger than it would be if the longitudinal tension from the 
steam pressure on the heads did not exist. 

If tension is applied to the ends of a rectangular prism, 
and external compression is added to all four sides, the true 
unit tension is much increased, or the piece is decidedly weaker 
in resisting the pull. 

Example. — At a certain point in a conical steel piston there 
exist principal stresses of 3,160 and 1,570 lbs,, of opposite signs. 
Then— _pi = 3,160, p^ = i ?i = 79°- V\ = i>570j V'x = 4/2 = 
390. ^1 -f /i = 3.550; ^, 4- /a — 2,360. 

Since test experiments to determine tensile and com- 
pressive strength are made by the application of a single 
direct force, the values so determined are compatible with the 



204 STRUCTURAL MECHANICS. 

existence of the opposite stress on planes at right angles with 
the cross-section. Hence the working stresses for any mate- 
rial may fairly be considered to be a little higher than ordinary 
-experiments show, provided account is taken at the same time 
•of all the stresses which act on a particle. 

208. Cooper's L#ines. — Steel plate as it comes from the 
inill has a firmly adhering but very brittle film of oxide of iron on 
the surface. This film is dislodged by the extension of a test spe- 
■cimen in tension when the yield point is passed. If a hole is 
punched at moderate speed in a steel plate, so that the particles 
under the punch have some opportunity to flow laterally under the 
compression, there will be a radial compressive stress in all direc- 
tions outwardly from the circumference of the hole. The unit 
■compressive stress will rapidly diminish as the circumference is 
left behind, and points will soon be reached where tension at right 
angles will be set up. If there is lateral crowding at points near 
the circumference there may be lateral compression. Presently at 
a certain distance the tension will be one-fourth the compression. 
Then, from the ellipse of stress, if jOg = 4 Pi and is of contrary 
sign, i CPi + P2) = fPi; and J {p, — p^) = § p„ and shearing 
planes will exist, lying through the points where a circle of radius 
^j9i cuts the ellipse. If _P2 is less than ^ p^ the shearing planes will 
lie nearer the direction of p^, and if p^ = p^, the shearing planes 
will make 45° with p^. The scale breaks on these lines of shear 
and there result curves where the bright metal shows through, 
branching out from the hole, intersecting and fading away. The 
process of shearing a bar will develop the same curves from the 
ilow of the metal on the face at the cut end. They are known as 
-Cooper's lines. 

These lines show that deformation takes place at con- 
siderable distances from the immediate point of shearing or 
punching. 

Examples. — i. A pull of 1,000 lbs. per sq. in. and a thrust of 
2,000 lbs. per sq. in. are principal stresses. Find the kind, direc- 
tion and magnitude of the stress on a plane at 45° with either prin- 
cipal plane. 

2. Find the stress per running unit of length of joint for a 
spiral riveted pipe when the line of rivets makes an angle of 45° 
with the axis of the pipe and when it makes an angle of 60°. 

0.707;); 0.5 p. 

3. A rivet is under the action of a shearing stress of 8,000 
lbs. per sq. in. and a tensile stress, due to the contraction of the 
rivet in cooling, of 6,000 lbs. per sq. in. Find jOj and j^o. 

j9j = — 11,540 lbs; p, = + 5,540 lbs. 



CHANGE OF FORM. 205 

4. A connecting plate to which several members are attached, 
as IV., Plate III., has a unit tension on a certain section of 6,500 
lbs. at an angle of 30° with the normal. On a plane at 60° with 
the first plane the unit stress of 5,000 lbs. compression is found at. 
45° with its plane. Find the principal unit stresses and the shear. 

— 6,600; -j- 4j8oo; 5,700. 

5. Assuming the weight of earth to be 105 lbs. per c. ft. and 
the horizontal pressure to be one-third the vertical, what is the 
direction and unit pressure per sq. ft. on a plane making an angle 
of 15° with the vertical at a point 12 ft. under ground, if the sur- 
face is level? 515 lbs.; 39^° with the horizon. 

6. A stand-pipe, 25 ft. diam., 100 ft. high. The tension in 
lowest ring, if J in. thick, is 7,440 lbs. per sq. in. If plates range 
regularly from -J in. thick at base to J in. at top, neglecting lap, 
the compression at base will be about 215 lbs. .per sq. in. For a 
wind pressure of 40 lbs. per sq. ft., reduced 50% for cylindrical 
surface, and treated as if acting on a vertical section, M at base = 
2,500,000 ft. lbs. Compression on leeward side at base =. 485 
lbs. per sq. in. If ^^ = — 7,440 lbs., 'P2 = 215 -}- 485 = 700 
lbs., find the stress and its inclination for a plane at 30° to the 
vertical? + 6,193 lbs.; 34° 40'. 

Prove that the shearing plane is 17° 04' from horizontal, and 
that the shear is 2,284 lbs. per sq. in. 



CHAPTER XII. 

rivets: pins. 

209. Riveted Joints. — There are four different ways in 
which riveted joints and connections may fail. The rivets 
may shear off; the hole may elongate and the plate cripple in 
the line of stress; the plate may tear along a series of rivet 
holes, more or less at right angles to the line of stress; or the 
metal may fracture between the rivet hole and the edge of the 
plate in the line of stress. From the consideration that a 
perfect joint is one offering equal resistance to each of these 
modes of failure, the proper proportions for the various riveted 
connections are deduced. 

210. Resistance to Shear. — The safe resistance of a 
rivet to shearing off depends upon the safe unit shear and 
the area of the rivet cross-section, which varies as the 
square of the diameter of the rivet. When one plate is drawn 
out from between two others, a rivet is sheared at two cross- 
sections at once, and is twice as effective in resisting any such 
action. Rivets so circumstanced are said to be in double 
shear, and their number is determined on that basis. 

211. Bearing Resistance. — The resistance against elon- 
gation of the hole or crippling the plate depends on the safe 
unit compression and what is known as the bearing area, the 
thickness of the plate multiplied by the semi-circumference of 
the hole. As the semi-circumference varies as the diameter, 
it is more convenient, and sufficiently accurate, to use the pro- 
duct of the thickness of the plate and the diameter of the 
rivet with a value of allowable unit compression about fifty 
per cent, greater than usual. 

212. Resistance of Plate. — The resistance to tearing 
across the plate through a line of holes, or in a zigzag through 
two lines of holes in the same approximate direction, depends 
on the safe unit tensile stress multiplied by the cross-section 



RIVETS: PINS. 207 

of the plate after deducting the holes. If the transverse pitch, 
or distance between centres of rivets, is considerable, an 
assumption of uniform distribution of tension on that cross- 
section is not likely to be true. 

The resistance of the metal between the rivet hole and 
the edge of the plate in the line of stress is usually taken as 
the safe unit shear for the plate multiplied by the thickness 
and twice the distance from the rivet hole to the edge. Some, 
however, consider that the resisting moment of the strip of 
metal in front of the rivet holes is called into action. 

213. Bending: Friction. — -There are those who advise 
the computing of a rivet shank as if it were subjected to a 
bending moment. If the rivet fills the hole and is well driven, 
there is no bending moment exerted on it, unless it passes 
through several plates. As practical tests have shown that 
rivets cannot surely be made to fill the holes, if the combined 
thickness of the plates exceeds five diameters of the rivet, 
this limitation will diminish the importance of the question of 
bending. 

No account is taken of the friction induced in the joint by 
its compression and the cooling of the rivet, and such friction 
gives added strength. As the rivet is closed up hot, the shank 
is under more or less tension when cold; the head is not given 
the .thickness required in the head of a bolt under tension. 
Therefore rivets are not available for any more tension, and 
should not be used for that purpose. Tight-fitting, turned 
bolts are required in such a case. 

214. Spacing. — The rivets should be well placed in a 
joint or connection, in order to insure a nearly uniform distri- 
bution of stress in the piece; they should be symmetrically 
arranged, be placed where they can be conveniently driven, 
and be spaced so that the holes can be definitely and easily 
located in laying out the work. See Plate III. 

215. Minimum Diameter of Rivets. — The punch must 
have a little clearance in the die. The wad of metal shears 
out below the punch with more ease and with less effect on the 
surrounding m.etal when it can flow, as it were, a little later- 
ally, and it then comes out as a smooth frustum of a cone 
with hollowed sides, reminding one of the vena contracta. 



2o8 STRUCTURAL MECHANICS. 

See Plate I. The punch must also be a little larger than the 
rivet, to permit the ready entrance of the rivet shank at a high 
heat. The diameter of the hole is often computed at | inch 
in excess of the nominal diameter of the rivet; but the rivet is 
treated as if of its nominal diameter. 

One other consideration has weight in determining the 
minimum diameter of the rivet. If the rivet is of less diame- 
ter than the thickness of the plate, the punch will not be 
likely to endure the work of punching. A diameter one and 
a half times the thickness of the plate is often thought 
desirable. 

216. Number and Size of Rivets. — Formulas are of 
little or no value in designing ordinary joints and connections. 
Boiler joints and similar work can be computed by formulas, 
but to no great advantage. Tables are used which give what 
is termed the shearing value of different rivet cross-sections 
in pounds, for a certain allowable unit shear, and the 'bearing 
or compression value of different thicknesses of plate and 
diameters of rivet, for a certain allowable unit compression. 
For a given thickness of plate, that diameter of rivet is the 
best whose two values, as above, most nearly agree. The 
quotient of the force to be transmitted through the connection 
or through a running foot of a boiler joint, divided by the less 
of -the two practicable values will give the minimum number 
of rivets. Their distribution is governed by the considerations 
previously referred to. Whether a joint in a boiler requires 
one, two or three rows of rivets depends upon the number 
needed per foot. 

Example. — Two tension bars, 6 in. by J in., carrying 30,000 
lbs., are to connected by a short plate on each side. Let unit 
shear be 7,500 lbs. per sq. in., unit compression 15,000 lbs., when 
diameter of rivet is used, and unit tension 12,000 lbs. The bear- 
ing value of T6 in. rivet in a I in. plate is 5,160 lbs., its shearing 
value in double shear is 2 X 2,780 = 5,560 lbs. A f rivet would 
give 4,700 and 4,600 respectively, but is of rather small diameter 
for thickness of plate. Hence 30,000 -^ 5,160 = 6 rivets neces- 
sary. If these rivets can be so arranged that a deduction of but 
one rivet hole is necessary from the cross-section of the tie, 
^6 — \^y^ =23! sq. in. net section, which will carry 31,125 lbs. 
at 12,000 lbs. unit tension. Each cover plate cannot be less than 
J in. thick, and, as will be seen presently, should be made a little 



rivets: pins. 



209 



more. The length will depend on the distribution of the rivets, 
to be taken up next. 

217. Arrangement of Rivets. — Long joints, under ten- 
sion, like those of boilers, are connected by one or more rows of 
rivets, as shown at A and B, Fig. 84. If more than one row 
is needed, the rivets are staggered, and the rows should be 
separated such a distance that fracture by tension is no more 
likely on a zigzag line than across a row. If the distance 
from centre to centre on the diagonal is not less than three- 



o 

D 
O 



B 




F 


/ \ 



000 


G 


000 

000 







o. 



o' 



,0 



■-O 








F.g. 84 



\ K 




(7) 
00 





fourths of the pitch, this object will be attained. To prevent 
tearing out at the edge of the plate, the usual specification of 
at least one and a half rivet diameters from centre oi hole to 
edge of plate will suffice. 

The tendency of a lap joint to cause an uneven distribu- 
tion of stress by reason of bending, and the same tendency 
when a single cover is used, is shown at C. The increase of 
stress thus caused should be offset by increased thickness of 
plate. An outside cover strip on the joint of a cylinder ex- 
posed to internal pressure is not so bad, as the internal 
pressure tends to force the ring to conform to a circle. A 



15 



210 STRUCTURAL MECHANICS. 

cover strip on either side is preferable, if not objectionable 
for other reasons. 

In splicing ties, D shows a bad arrangement, the upper 
plan failing to distribute the stress evenly across the tie, and 
the lower plan wasting the section by excessive cutting away. 
The rivets at E are* well distributed across the breadth, and 
weaken the tie by but one hole, as only two-thirds of the stress 
passes the section reduced by two holes; and, unless the net 
section at this place is less than two-thirds of the section 
reduced by one hole, it is equally strong. Thus b — ^ = 
\{b — 2d), or a breadth equal to or greater than four diameters 
will satisfy this requirement. The covers, however, will be 
weakened by two holes, and hence their combined thick- 
ness, when two are used, should exceed the thickness of 
the tie. 

F similarly is better than G, and the tie at F is again 
weakened by but one hole. To prevent the great weakening 
of the cover in this case the rivets may be spaced as shown at 
H; but it is doubtful whether the saving in thickness of the 
covers is not offset by the increase in length. 

As it is desirable to transmit all but the proper fraction of 
the tension past the first rivet, the corners of the cover F or H 
are clipped off, thus increasing the unit tension in the reduced 
section and increasing its stretch to more nearly correspond 
with the unit tension and elongation of the tie beneath. The 
appearance of the connection is also improved. 

218. Remarks. — If the member is in compression, the 
holes are not deducted, since the rivets completely fill the 
holes; and the strength is computed on the gross section. 
Unless special care is exercised in bringing two connected 
compression pieces into close contact at their ends, good 
practice requires the use of a sufficient number of rivets at 
the connection to transmit the given force. 

Rivet heads in boiler work are flat cones. In bridge and 
structural work they are segments of spheres, known as button 
heads, and are finished neatly by means of a die. These 
heads may be flattened when room is wanting, and counter- 
sunk heads are used where it is necessary to have a finished 
flat surface. 



rivets: pins. 211 

Members which meet at an angle, are connected by plates 
and rivets. See Plate III. The axes of the several members 
should if possible intersect in a common point. If they do 
not, moments are introduced which give rise to what are 
known as secondary stresses, as distinguished from the primary 
stresses due to the direct forces in the pieces of the frame. 
Such secondary stresses may be of considerable magnitude in 
an ill-designed joint. 

It is desirable to arrange the rivets in rows which can be 
easily laid out in the shop, and a central rivet, where several axes 
of pieces intersect, furnishes a convenient point of reference. 

Commercial rivet diameters vary by sixteenths of an inch, 
more commonly by eighths, |, |, | and one inch beingtheones 
frequently used. As much uniformity as possible in the size 
of rivets will tend to economy in cost. 

2ig. Structural Riveting. — The following rules for 
structural work are in harmony with good practice: — 

Holes in steel | inch thick or less may be punched; when 
steel of greater thickness is used, the holes shall be drilled. 

Rivets shall have round concentric heads, of a depth at 
the circumference of the shank of not less than one-half the 
diameter of the rivet, and with full bearing on the plate. 

The pitch of rivets, in the direction of the stress, shall 
never exceed 6 inches, nor 16 times the thickness of the thin- 
nest plate connected, and not more than 30 times that thick- 
ness at right angles to the stress. 

At the ends of compression members the pitch shall not 
exceed 4 diameters of the rivet, for a length equal to twice 
the width of the member. 

The distance from the edge of any piece to the centre of 
a rivet hole must not be less than ij times the diameter of the 
rivet, nor exceed 8 times the thickness of the plate; and the 
distance between centres of rivet holes shall not be less than 
3 diameters of the rivet. 

The diameter of the die shall not exceed that of the 
punch by more than tV of an inch, and all rivet holes shall be 
so accurately spaced and punched that, when the several parts 
are assembled together, a rivet rV inch less in diameter than 
the hole can generally be entered hot into any hole. 



2 12 STRUCTURAL MECHANICS. 

The effective diameter of a driven rivet will be assumed 
to be the same as its diameter before driving; but the rivet 
hole will be assumed to be one-eighth inch diameter greater 
than the undriven rivet. 

220. Boiler Riveting, — The following proportions are 
sometimes used for tanks, stand-pipes, and similar work: — 

Diameter of rivet, double the thickness of the plate. 
Pitch, centre to centre, 3 diameters of the rivet for a single 
row; 4 diameters for double or triple rows, with rivets stag- 
gered (zigzag) 3 diameters on the diagonal line. From centre 
of rivet line to the edge of plate, after it has been beveled to 
60° for calking, i^ diameters + \ inch. 

Unwin gives the following rules for riveted joints: — 

Single riveted lap joints: — Diameter of rivet, twice the thick- 
ness of plate; pitch of rivets and width of lap, three times the 
diameter of rivet. Butt joints with single cover: — The same as 
above. 

Double riveted lap joints: — Diameter of rivet, twice the thick- 
ness of plate; pitch of rivets, four and a half times the diameter of 
the rivet; width of lap six diameters in zigzag riveting. 

Butt joints with double covers, each cover being one-half the 
thickness of plate: — Diameter of rivet, one and a half times thick- 
ness of plate; pitch in single riveted joints, 3 J diameters, and 
width of cover strips, 6 diameters; pitch in double riveted joints, 
5 diameters, and width of cover strips, 12 diameters in zigzag 
'riveting. 

In ordinary cases there is no danger that the rivets will 
be too far apart to render the joint water or steam tight, when 
the edge of the plate on one or both sides is properly closed 
down with a calking tool. 

221. Strength of Joint. — Rivet steel may be required to 
have an ultimate strength of from 54,000 to 62,000 lbs., a 
yield point of 30,000 lbs., and an elongation of 26 per cent. 
Similarly, rivet iron may show 50,000, 26,000, and 18 per 
cent. 

The strength of a well-designed, single riveted joint may 
be 50 per cent.; of a double riveted joint, 65 per cent.; and 
of a triple riveted joint, 80 per cent, of that of the unpunched 
plate. 

For unit stresses for shear and bearing see § 177. 



RIVETS: PINS. 213 

222. Pins: Reinforcing Plates. — The pieces of a frame 
are frequently connected by pins instead of rivets. The axes 
of the several pieces are thus made to meet in a common 
point, if the pin hole is central in each member. Pins are 
subjected to compression on their cylindrical surfaces, to shear 
on the cross-section, and to bending moments. The com- 
pression on the pin-hole is reduced to the proper unit stress, 
if necessary, by riveting reinforcing plates to the sides of the 
members, as shown at K, Fig. 84. A sufficient number of 
rivets to transmit the proper proportion of the force must be 
used, with a due consideration of the shearing value of a rivet 
and its bearing value in the reinforcing plate or the member 
itself, which ever gives the less value. No more rivets should 
be considered as efficient behind the pin than the section of 
the reinforcing plate each side of the pin hole will be equiva- 
lent to. 

When the pin passes through the web of a large built 
member such as a post or a top chord of a bridge, the web is 
often so thin that more than one reinforcing plate on either 
side is needed. It is then economical to make the several 
plates of increasing length, the shortest on the outside, and 
determine the number of rivets in each portion accordingly. 
The longest plate in such a case is sometimes required to 
extend in front of the pin four times the transverse distance 
from the pin centre to the line of rivets in the angle iron, so 
that the stress may be transferred to the flange angles and 
plate, and not overtax the web. 

223. Shear and Bearing. — The shear at any section of 
the pin is found from the given forces in the pieces connected. 
The resultant of the forces in the pieces on one side of any 
pin section will be the shear at that section. As the pin will 
probably not fit the hole tightly (a difference of diameter of 
one-fiftieth of an inch being usually permitted), the maximum 
unit shear will be four-thirds of the mean, § 86. Specifica- 
tions frequently give a reduced value for mean unit shear, 
which provides for this unequal distribution. 

Bearing area is also figured as if projected on the diame- 
ter with, e. g., 15,000 lbs. in place of 10, 000 lbs. per sq. inch 
on the semi-circumference. 



2i4 



STRUCTURAL MECHANICS. 




224. Bending Moments on Pins. — At a joint where 
several pieces are assembled, the resisting moment, required 
to balance the maximum bending moment on the pin caused 
by the forces in those pieces, will generally determine the 
diameter of the pin. In computing the bending moments, 
the centre line of each piece or bearing is considered the point 
of application of the force which it carries. This assumption 

is likely to give a result somewhat in 
excess of the truth, as any yielding 
tends to diminish the arm of each 
force. 

The process of finding the bend- 
ing moments will be made clear by an 
illustration. Fig. 84, A, shows the 
plan and elevation of the pieces on 
3 a pin, with the forces and directions 
marked. The thickness of the pieces, 
3 which are supposed to be in contact, 
is also shown. The joint must be 
symmetrically arranged, to avoid tor- 
sion, and simultaneous forces must be used, which reduce to 
zero for equilibrium. As the joint is symmetrical, the com- 
putation is carried no farther than the piece adjoining the 
middle. ' 

Resolve the given forces on two convenient rectangular 
axes, here horizontal and vertical. Set the horizontal com- 
ponents in order in the column marked H, the vertical ones 
in the column marked V. Their addition in succession gives 
the shears, marked F. The next column shows the distance 
from centre to centre of each piece. Ydx is then the incre- 
ment of bending moment; and the summation of increments 
gives, in the column M, the bending moment at the middle 
of each piece, from the horizontal and from the vertical com- 
ponents respectively. The square root of the sum of the 
squares of any pair of component bending moments will be 
the resultant bending moment at that section. It is compara- 
tively easy to pick out the pair of components which will give 
a maximum bending moment on the pin. Equate this value 



Fig. 84 A. 



rivets: pins. 215 

with the resisting moment of a circular section and find the 
necessary diameter. 

H. F. dx. Ydx, M. 
A -\- 10,000 -j- 10,000 

B 40,000 30,000 -f" II;250 -j- 11,250 

C o — 30,000 — 22,500 — 11,250 

D -|- 15,000 — 15,000 — 18,750 — 30,000 

E +15,000 o —13,125 —43,125 












, 






V. 










A 


— 5,780 


— 5,780 


i>^ 






B 





— 5,780 


^ 


6,503 


6,503 


C 


— 2,890 


8,670 


/8 


— 4,335 


— 10,838 


D 





8,670 


lA 


5,419 


— 16,257 


E 


+ 8,670 





/8 


— 7,586 


— ^Z^'^AZ 



M at D = 1/(30,0002 _!_ 16,2572); M at E = -/(43,i252 + 
23,843^). The latter is plainly the larger, and is 49,210 in. lbs. 

The pieces can be rearranged on this pin to give a smaller 
moment. The maximum moment is not always found at the 
middle. 

The bending moment at any point of the beam or shaft, 
when the forces do not lie in one plane, can be found in the 
same way. 

A solution of the above problem by graphics can be 
found in the author's Graphics, Part 11. , Bridge Trusses. 

For values of unit stress in pins, see § 177. 

For bolts, see § 134. 

Examples.— \. A tie bar, ^ in. thick, and carrying 24,000 
lbs., is spliced with a butt joint and two covers. If unit shear is 
7,500 lbs., unit bearing on diam. is 15,000 lbs., and unit tension 
is 10,000 lbs., find the number, pitch and arrangement of 3^ in. 
rivets needed, and the width of the bar. 



2l6 STRUCTURAL MECHANICS. 

2. The longitudinal lap joint of a boiler must resist 52,000 
lbs. tension per linear ft. If the unit working stress for the shell 
is 12,000 lbs. and the other stresses as above, what size of rivet is 
best, for double riveting, what the pitch, and the thickness of 
the shell? 

3. A pin at VIII., Plate III., is to be computed. The force 
in the top chord at the right is 120,000 lbs., at the left 80,000 lbs., 
in the post 40,000 lbs., and the horizontal component of the ten- 
sion in main tie, which slopes at 45°, is 40,000 lbs. The vertical 
plates of the chord are 12 in. apart, the two ties 9^ in., and the 
side plates of the post, 8 in. If/ for bending is 20,000, what is 
the diameter of the pin ? Will the plates need reinforcing, if j^ 
in thick? 



CHAPTER XIII. 



ENVELOPES, 



225. Stress in a Thin Cylinder. — Boilers, tanks and 
pipes under uniform internal normal pressure of / per. sq. 
inch. 

Conceive a thin cylinder, of radius r,, to be cut by any 
diametral plane, such as the one represented in Fig. 85, and 
consider the equilibrium of the half cylinder, which is illus- 
trated on the left. It is evident that, for unity of distance 
along the cylinder, the 
total pressure on the di- 
ameter, 2 pr, must balance 
the sum of the components 
of the pressure on the 
semi-circumference in a 
direction perpendicular to 
the diameter. This pressure 2 pr, uniformly distributed over 
the diameter, must cause a tension T in the material at each 
end to hold the diameter in place. Hence 




T = 'pr. 

As all points of the circle are similarly situated, the tension in 
the ring at all points is constant, and equal to pr. If the 
thickness is multiplied by the safe working tension /per square 
inch, it may be equated with pr, giving. 



Required net thickness = ^r -^ f. 

In a boiler or similar cylinder made up of plates an increase 
of thickness will be required to compensate for the rivet holes. 
If a is the pitch, or distance from centre to centre, of consecu- 
tive rivets in one row along a joint, and d' the diameter of the 
rivet hole, the effective length a to carry the tension is reduced 



2l8 STRUCTURAL MECHANICS. 

to a — d', and the gross thickness of plate must not be less 
than <_. 



/ a — d' 

Example.— T\i^ circumferential tension in a boiler, 4 ft. 

diam., carrying 120 lbs. steam pressure is 120 . 24 = 2,880 lbs. 

per liiiear inch of length of shell, which will require a plate 

2,880 . , . T . , . , ^ . , 

m. thick (net), it / is not to exceed 10,000 lbs. per sq. m. 

10,000 ^ 

Net thickness = i-i in. If a longitudinal joint has fin. rivet holes, 

at 2i in. pitch, in two rows, the thickness of plate must not be less 

2,880 . 2I 

than = 16 m. 

10,000 . li 

226. Another Proof of the value of T may be obtained as 
follows: — The small force on arc ds = 'pds. The vertical component 

of this force = 'pds sin d = pdx. The entire component on one side 

/+ r 
pdx = 2 pr, which must be resisted by the 

tension in the material at the two ends of the diameter. 

The same result will be obtained graphically by laying off a 
load line = 2 pds, which becomes a regular polygon of an infinite 
number of sides, /. e., a circle, with the lines to the pole making 
the radii of the length pr. 

The cylinder, under these circumstances, is in stable 
equilibrium. If not perfectly circular, it tends to become so, 
small bending moments arising where deviation from the circle 
exists. Hence a lap joint in the boiler shell causes a stress 
from the resisting moment to be combined with the tension at 
the joint. 

The above investigation applies only to cylinders so thin 
that the tension may be considered as distributed uniformly 
over the section of the plate. 

For riveting see Chapter XII. 

227. Stress in a Right Section. — The total pressure 
from / on a right section of the cylinder is 7rr^ p, which will 
also be the resultant pressure on the head in the direction of 
the axis of the cylinder, whether the head is flat or not. This 
pressure causes tension in every longitudinal element of the 
cylinder, or in every cross-section. As this cross-section is 
2 7rr X thickness, the longitudinal tension per linear inch of a 
circumferential joint is tt^^ / ^ 2 ^r = J /r, or one-half the 



ENVELOPES. ^19 

amount per linear inch of a longitudinal joint. Hence a 
boiler is twice as strong against rupture circumferentially as 
longitudinally. Hence, also, the longitudinal seams are often 
double riveted, while the circumferential ones are single 
riveted. 

228. Stress in any Curved Ring under Normal Press- 
ure. — Since the stress in a circular ring of radius r, under internal 
or external normal unit pressure p, is 'pr per linear unit of cross- 
section of the ring, and per sq. inch is pr -^ thickness, being ten- 
sion in the first case and compression in the second case, the direct 
stress on the cross-section of any curved ring, at a point where the 
applied pressure is normal, will be given by the same expression, 
if for r is substituted the radius of curvature of the ring at that 
point. Unless the ring, however, has the form of equilibrium 
under the given applied forces, a resisting moment also is required 
at the cross-section; but the resultant force there may be found as 
stated. 

Example. — An elliptic ring, diameters 60 in. and 30 in., has 
a normal pressure exerted on it, at the extremity of the shorter 
diameter, of 150 lbs. per linear inch of ring. The radius of curva- 
ture at that point is or' -=- b, or 30'^ -^ 15, and the resultant force 
at that section will be 150 . 30^ -^ 15 = 9,000 lbs. 

229. Thin Spherical Shell: Segmental Head. — If a 

thin hollow sphere of radius r' has a uniform normal unit 
pressure / applied to it within, the total interior pressure on a 
meridian plane will be ^/^ /, and the tension per linear inch 
of shell will be 

Ttr'"^ p -^ 2-r' = \p^'- 

If / is applied externally, the stress in the material will be 
compression. It may be noted that the double curvature of 
the sphere is associated with half \\\q^ stress which is found in 
the cylinder of single curvature having the same radius. 

If a segment of a sphere is used to close or cap the end 
of a cylinder or boiler, the same value will hold good. In this 
case the radius / is greater than r for the cylinder. 

If the segmental end is fastened to the cylinder by a 
bolted flange, the combined tension on the bolts will be "r^p, 
as this is the total force on a right section of the cylinder. 

The flange itself will be in compression. The pressure 
/ from below, in Fig. 86, causes a pull per circumferential 
unit, in the direction of a tangent at B, which pull has just 




2 20 STRUCTURAL MECHANICS. 

been shown to be equal to ^pr': It may be resolved into 
vertical and horizontal components. The vertical component 
B C is, by § 227, \pi'. The horizontal component h must be 
proportioned to the vertical component as A O to A B, the 

sides of the right angled triangle to 
which they are respectively perpen- 
dicular. As A O = i/(r'' — r'), 
h : ^pr = ■,/(r'^ — r^) : r, 
or /^ = ^p-^{r''' —r'). 

As h is a uniform normal pressure 
applied from without (or tension 
applied from within) in the plane of the flange, the com- 
pression on the cross-section of the latter will be hr or 
i/r-j/(r'^ — ^^^)» to be divided by that cross-section for find- 
ing the unit compression. 

Segmental bottoms of cylinders are sometimes turned 
inward. The principles are the same. 

Example. — A segmental spherical top to a cylinder of 24 in. 
diam., under, 100 lbs. steam pressure, has a radius of 15 in. with 
a versed sine of 6 in. The tension in top = \ . 100 . 15 = 750 
lbs. per linear inch. If its thickness is \ inch, the stress per sq. 
in. is 3,000 lbs. The total pull on the flange bolts is 100 . 144 . 22 
^ 7 = 45,260 lbs. A f in. bolt has about 0.3 sq. in. section at 
bottom of thread, giving a tension value of about 3,000 lbs. if 
y" = 10,000 lbs. There would be needed some 15 bolts, about 5J 
in. centre to centre on a circumference of 26 in. diameter. The 
compression in the flange is \ . 100 .12.9 = 5,400 lbs. A 2 in. 
by J in. flange, with a f in. hole has a section ^ . ij = |- sq. in., 
giving a unit compression in the flange of f . 5,400 = 8,600 lbs. 
per sq. in. 

A similar compression acts in the connecting circle between 
a water tank and the conical or spherical bottom sometimes built. 
See §§ 239, 240. 

230. Collapsing of Tubes. — If a uniform normal pres- 
sure acts on a thin hollow cylinder from without, any ring is 
in unstable equilibrium, and any slight deviation from the 
circular form develops a bending moment, equal to /r multi- 
plied by the deviation ordinate, which bending moment must 
be resisted by the ring. If the cylinder is quite thin, it has 
little ability to resist such a moment, and the cylinder or tube 
is in danger of collapsing. But, manifestly, if the cylinder is 



ENVELOPES. 221 

closed at its ends, or is reinforced by rings or by internal dia- 
phragms at intervals, the movement inwards at any point 
develops a resisting moment in the longitudinal elements. 
Hence a short, closed cylinder, or one with rings or flanges at 
proper intervals will be prevented from collapsing. The ex- 
perimental relation between the pressure / which will cause 
collapsing, the length /, thickness t and diameter d of an iron 
tube, is given by Fairbairn as 

p ^ 9,672,000 f -^ /^nearly, 

which appears to be, p ^r. Yjf -^- 3/^, or / 1= Yjf" -f- 3/^, all in 
inches. 

For safety, / should be very much less. 

The collapsing of an empty water tower or stand-pipe, 
under a strong wind pressure, is due to similar action, although, 
as the wind pressure is exerted on but one side, and is by no 
means uniformly distributed on the semicircle, the tendency 
to collapse is far greater. It is guarded against by one or 
more angle-iron rings riveted at or near the top. A 3 in. by 
5 in. angle at the top will suffice for a tank 20 ft. in diameter; 
two similar angles for one 30 ft. in diameter; and bracing like 
a bicycle wheel will hold the top of larger tanks. 

231. Stand-Pipes. — The following data for stand-pipes 
and water tanks are in keeping with good practice: — Use 
57,000 to 65,000 lb. steel plates, showing 20 per cent, 
elongation in 8 inches. Allow \ inch for corrosion, and use no 
plates thinner than \ inch. The bottom of tanks may be ^or 
T6 inch thick, unless greater strength is required from form of 
bottom or means of support; the bottom of stand-pipes may 
be I or J inch thick. Rivet heads should not be countersunk 
on the bottom, as full heads make tighter work. Bed the 
bottom on a fresh cement grouting on top of foundation 
Plates usually build 5 ft. rings. Alternate inside and outside 
rings are best. Internal angle-iron should be used at bottom, 
calked on both edges. If more bearing is wanted for the 
shell in large pipes, add outside angle. For riveting, 
see § 220. 

See also, Engineering Record, Feb. 11, 1893. 



222 STRUCTURAL MECHANICS. 

Example. — Detailed calculation for riveting. A standpipe, 
loo ft. high, 25 ft. diameter, full of water at 62^ lbs. per c. ft. 
Let/" =r 12,000 lbs. tension and shear, and 10,000 lbs. compres- 
sion, that is, 15,000 lbs. on diameter of hole, per sq. in. The ten- 
sion per linear foot of the lowest ring will be 62 J . 100 . i2|- = 
78,125 lbs. 78,125 lbs. -^ 12,000 = 6. 51 sq. in. of iron required 
per linear foot. As the final thickness will much exceed \ in., try 
|, ^ and I in. rivets. 

f in., area = 0.442; shearing value -— 5,300 lbs.; 15 needed per ft. 
I " " 0.601 " " 7,220 II " " " 

I " " 0.758 " " 9,430 8i " " " 

If 3 rows of \ in, rivets are used, 5 in a row, the pitch will be 
2^ in. The hole being called |^ in. diameter, 5 x J = 4| in. 
12 — 4f = 7f in., nef width of i ft. of sheet. 6.51 -^ 7.62 = 
0.854 or J in., required thickness of plate, f • J • 15,000 = 9,844 
lbs., bearing value of i rivet. There are too many rivets for bear- 
ing, and the plate is too thick for \ in rivets. 

If 3 rows of ^ rivets are tried, 3§ in a row per ft., the pitch 
will be 2i\ i^* Hole is i in. diameter. 12 — ^y^ ^ ^\ in. net 
width. 6.51 -^ 8.33 = 0.78 in. thickness of plate. A \ in. plate 
will serve, as pressure decreases upward on the joint, and the 
allowance for the hole is large. Eight rivets only are needed for 
bearing. If the distance between rows is made 2\ in., the lap will 
be 8 in., or, for a cover strip, 16 in. 

If I in. rivets are used, 3 rows with 4J in. pitch will be 
required, with \ in. plate, giving a bearing value per ft. of joint 
of 95,600 lbs. 

So much detail is not necessary after a little experience. 

232. Thick Hollow Cylinder. — If the walls of a hollow 
cylinder or sphere are comparatively thick, it will not be suf- 
ficiently accurate to assume that the stress in any section is 
uniformly distributed throughout it. If the material were 
perfectly rigid, the internal or external pressure would be 
resisted by the immediate layer against which the pressure 
was exerted, and the remainder of the material would be 
useless. As, however, the substance of which the wall is 
composed yields under the force applied, the pressure is trans- 
mitted from particle to particle, decreasing as it is transmitted, 
since each layer resists or neutralizes a portion of the normal 
pressure, and undergoes extension or compression in so doing. 

233. Greater Pressure on Inside. — Let Fig. %"] repre- 
sent the right section of a thick, hollow cylinder, such as that 
of an hydraulic press. Let i\ and r.^ be the internal and 



ENVELOPES. 



223 



external radii in inches; px and /g the internal and external 
normal unit pressures in pounds per square inch, p^ being the 
greater; and p the. unit normal pressure on any ring whose 
radius is r. 

If a hoop is shrunk on to the cylinder, p^ will be the unit 
normal pressure thus applied to the exterior of the cylinder. 

The unit tensile stress found in a thin layer of radius r 
and thickness dr will be denoted by t, and will be due to that 
portion of p which is resisted by the layer and not trans- 
mitted to the next exterior layer. As the tension in a ring of 
radius r, under any interior normal unit pressure / is pr, the 
entire tension on a section from rj to r must be p^r^ — pr, 

/r 
tdr. As / and r are vari- 
n 
ables, there is obtained by differentiating the equation 



p\n — Pr 



/r 
tdr, 
r\ 



— d i^pr) = tdr, 
or pdr -{- rdp -\- tdr = o. 



('•) 




Another equation can be deduced from 
the enlargement of the cylinder. The fibres 
or- layers between the limits ri and r, being 
compressed, will be diminished in thick- 
ness. The compression of a piece an 
inch in thickness by a unit stress / will be 
p -^ K, § 10, and of one dr thick will be 
pdr -^ E. The total diminution of thickness between 7\ and 

I f^ 
r, from what it was at first, will therefore he ^=r pdr. 

^ J ri 

But the annular fibre or ring whose radius is r, and 

length 2-r, has been elongated / -^ E per inch of length. 
Its length will now be 2-r ( 1-)--=-) and its radius r ( i-j ] . 

The internal radius must similarly have become r^ ( i-f ^ | , 
where / is the value of t for radius r^ . The thickness r — r^ 
has now become ^' ( i + -r^ j — ^'i [^ ~^ 4^) > ^^^> ^Y ^ub- 



2 24 STRUCTURAL MECHANICS. 

tracting this value from r — r^ , there is found the diminution 

of thickness, i\ -.^ '' ^ • This expression may be equated 

with the previous one for decrease of thickness, or 



ft 



"r 



'-, ^ - '^ ^ = ^ / Pdr. 



n 



Since the first term is constant, there is now obtained by 
differentiating this equation, 

— d[tr^ =. pdr, or tdr -f- rdt -f- pdr = o. (2.) 

Add (i.) and (2. )., and multiply by r to make a complete 
differential. Then integrate. 

2(/ -\- p)rdr -f r'(^/ -\- dp) — o 
r\t^p) = Constant; .- . =. r,\f ^ p,) ^ r,%/' + /,)• (3-) 

Again: subtract (i.) from (2.), and then integrate, 

d/ — dp ^^ o. t — / = Constant; . • . = / — /j =.: /' — p<^. (4.) 

From (3. ) and (4. ) are obtained, by addition and sub- 
traction, 

If the internal radius is given, the external radius, and 
hence the required thickness, r^ — i\, is found by eliminating 
f from (3.) and (4. ) 

'. - '. vij^^^y (..> 

If /2 is atmospheric pressure, it may be neglected when 
/>i is large. In that case 



As ^2 becomes infinite when the denominator of (6.) is 
zero, it appears that no thickness will suffice to bring y within 
the safe unit stress, if /i exceeds/" + 2/2- 

These formulas do not apply to bursting pressures, nor to 
those which bring /"above the elastic limit; for E will not then 



ENVELOPES. 225 

be constant. They serve for designing or testing safe con- 
struction. 

Examples. — Cylinders of the hydraulic jacks, for forcing for- 
ward the shield used in constructing the Port Huron tunnel, were 
of cast-steel, 12 in. outside diam., 8 in. diam. of piston, with Jin. 
clearance around same; pressure 2,000 lbs. per sq. in. 

r} 36 . 16 /+ 2,000 

— = Q — = 7 • / = 6,030. 

r{ 289 / — 2,000 

A cast-iron water-pipe, at the Comstock mine, was 6 in. bore,. 
2 J in, thick, and was under a water pressure of 1,500 lbs. on the 
sq. in., or aboat 3,400 ft. of water. Here/= 2,770 lbs. per sq. 
in. for static pressure, while the formula for a thin cylinder gives 
1,800 lbs. 

234. Greater Pressure on Outside. — In this case the 
direction or sign of t will be reversed, it being compression in 
place of tension. From the preceding equations, without 
independent analysis, by making t negative, there result, 

— d(^pr) — — tdr) d{tr) = pdr, 
pdr -\- rdp — tdr = o; tdr -]- rdt — pdr = o. 
r\p - /) = r:\p, -/) =. r/(A -/')• 

t -h/ =/ + /i =/' +A. 

The outer radius and pressure will now be taken as given 
quantities, and the unit compression in the ring at any point 
will be 



, _ /' + ?>. , >± r-p. . „ _ f' +ih r} r ~v. 

2 r- 2 2 r- 2 



(7-) 



. = wC-^:^^). (8.) 

which becomes, if p^ is neglected as small, 

The external pressure j^o must be less than ^{f -\- Pi), if 
Ti is to have any value. It will be seen from ^ in (7. ) that the 

compression is greatest at the interior. 

16 



2 26 STRUCTURAL MECHANICS. 

Example. — An iron cylinder, 3 ft. internal diameter, resists 
1,150 lbs. per sq. in. external pressure. The required thickness, 
\i f ^=. 9,000 lbs , is given by 



8 = r, ,/('i - M^V 0.86; 

V Q.OOO / 



9,000 / ^' 

r^ = 20.9 in. Thickness = 3 in. 

235. Action of Hoops. — To counteract in a greater or 
less degree the unequal distribution of the tension in thick, 
hollow cylinders for withstanding great internal pressures, 
hoops are shrunk on to the cylinders, sometimes one on 
another, so that, before the internal pressure is applied, the 
internal cylinder is in' a state of circumferential compression, 
and the exterior hoop in a state of tension. If the internal 
pressure on the hoop is computed, for a given value of /in the 
hoop, and this pressure is then used for w on the cylinder, the 
allowable internal pressure pi on the cylinder consistent with 
;a permissible /in this cylinder can be found. There is, how- 
ever, an uncertainty as to the pressure 2^2 exerted by the hoop. 

Examples. — A hoop one inch thick is shrunk on a cylinder of 
6 in. external radius and 3 in. internal radius, so that the max. 
unit tension in the hoop is 10,000 lbs. per sq. in. This stress, by 
§ 233, will be due to an internal pressure on the hoop of 1,530 lbs. 
per sq. in. 

For 7 = 6 A]22221±ll\ or 19 = '°'°°° + ?>.. 
viOjOoo — 'p^J 36 10,000 J)^ 

This external pressure p^ on the cylinder will cause a compressive 

unit stress in the interior circumference of the cylinder when 

■empty, after the hoop is shrunk on, of 4,080 lbs., and will permit 

,an internal pressure in the bore of 8,448 lbs. per sq. in., consistent 

36 10,000 -I- p, 

•w'ith / = 10,000 lbs. For — — -^ — . 7'he cyl- 

9 10,000 — I?! -f- 3,000 

inder alone, without the hoop, would allow a value of j)^ given by 

-26 10,000 + P, , 1, TT 1 1- 1 1 1 T 

'^— = —, 01* Pi = 0,000 lbs. It the cylinder had been 

9 10,000 — pi 

4 in. thick, the internal pressure might have been 6,900 lbs. The 

gain with the hoop, for the same quantity of material, is 1,548 lbs., 

or some 22 per cent. 

Hydraulic cylinder for a canal lift at La Louviere, Belgium, 

6 ft. 9 in. interior diam., 4 in. thick, of cast-iron, hooped with 

steel. Hoops 2 in. thick, and continuous. When tested, before 

hooping, one burst with an internal pressure of 2,175 ^^s. per sq. 



ENVELOPES. 227 

in., one at 2,280 lbs., and a third at 2,190 lbs. These results, if 
the formula is supposed to apply at rupture, give an average ten- 
sile strength of 23,400 lbs. per sq. in. The hoops were supposed 
to have such shrinkage that an internal pressure of 540 lbs. per sq. 
in. would give a tension on the cast-iron of 1,400 lbs., and on the 
steel of 10,600 lbs. per sq. in. The ram is 6 ft. 6J in. diam., and 
3 in. thick, of cast-iron, an example of the greater pressure outside. 
236. Thick Hollow Sphere. — Greater pressure on inside. 
Let Fig. 87 represent a meridian section of the sphere. Suppose 
f, t, etc., to be perpendicular to the plane of the paper. The 
entire normal pressure on the circle of radius 7\ will be p^ ~^'i, and 
the tension on the ring between radii r^ and ;- will be - (/j ;-/- — P^'')- 
Any ling of radius r and thickness dr will carry 2 -;-/ dr, and hence 
is derived the first equation 

r 

~ (Pi ^'\ — P^'^) = 2 - rt dr, or — d {p7'^) ~ 2 r/ d?'. 

J '\ 
r- dp -\- 2 pr di^ -j- 2 7't dr = o. 

The second equation will be the same as obtained for the cylinder, 

— d (/r) = pdr, or rdt -|- tdr -}- pdr = o. 

Strike out the common factor r from the first equation, multiply 
the second by 2, and subtract. 

2 rdt — rdp = 0, or 2 dt — dp ^ o. 
2 t — / = Constant; .-. = 2/ — p^ = 2f' — p^_. (9.) 

Again: add the first to the second and multiply by r". 

r' {dp 4- dt) + 3 1"' dr {p -\- t) = o. .-. 
r' (/ + - Constant; .-. = r,' {f -\- p,) = r/ (/+/,). (10.) 

From (9.) and (lo.), 

3 '' 3 3 '^ 3 

These formulas are not applicable to bursting pressures for 
the reason given before. For a finite value of r.^, p^ must be less 
than 2 /-j- 3 /2- I^ A is atmospheric pressure, it may be neglected, 

and 



228 



STRUCTURAL MECHANICS. 



237. Sphere: Greater Pressure on Outside. — Here 
again / changes to compression or reverses in sign, yielding 



/ = i/:i-^+ 



/'-A 



'1 - ^2x/( 



/ 



2/+A 

3 



f — Pi 



^ f + Pi — z P: 



) 



(I3-) 



2 (/-A) 

That ^1 shall be greater than zero requires that /a < J (2/ + /1). 

238. Diagrams of Stress.— Curves may be drawn to 
represent the variation of / and / in the four preceding cases. 
They are all hyperbolic, and, if r is laid off from the centre O 
on the horizontal axis, each curve will have the vertical axis 
through O for one asymptote, and for the other a line parallel 
to the horizontal axis, at a distance indicated by the first term 
in each value of t or /. The four accompanying sketches 
show the various curves. The values of /and/', the unit 
stresses in the material at the interior and exterior, which cor- 
respond to the given 




F'.g 89 



'/z(j*y,) \ 




values of p^ and p^y 
are found at the 
extremities of the ab- 
scissas which repre- 
sent Ti and To. The 
error which would 
arise from consider- 
ing / as uniformly dis- 
tributed is manifest. 
The dotted circles 
show the respective 
cylinders or spheres. 
Fig. 88 gives the ex- 
ternal and internal 
tensile stress for p^ in 
the interior of a thick cylinder. Fig. 89 shows the distribution 
of compression when the greater pressure is from without. 
Figs. 90 and 91 represent thick spheres under similar pressures. 
239. Tank with Conical Bottom. — A water tank of 
radius r may be built with a conical bottom and be supported 
at the perimeter only. Let the angle subtended by the cone 





ENVELOPES. 



229 




Ftg. 9a. 



at the vertex be 2 e, the distance along any element from the 
vertex D, Fig. 92, to any point P be a, and the depth of water 
above the vertex be h. Then the normal pressure at P, if w 
is the weight of a cubic unit of water, will be w {h — a cos d). 
If a plane of section is passed 
perpendicularly to the element 
D C at P, it will cut from the 
cone a conic section, usually an 
hyperbola. The radius of curva- 
ture, p, of that curve at its vertex P 
can be proved to be P I = rt: tan d. ^ 
From the fact pointed out in 
§ 228, that the stress, at a point 
in a curve where the external 
pressure is normal, is equal to the 
product of the unit pressure and "-'^ 

the radius of curvature, — the ten- 

sio7i at P, per unit of length of a radial joint along the 
element D C, will be 

pp =120(^/1 — a cos 6) a tan = wa (Ji tan 6 — a sin 6"). 

As h is usually longer than D C, this tension will increase 
from D to C, being zero at D, and at C being equal to 

7£/ . D C {h tan ^ — D C sin ^) — lu . \) C {h tan 6 — r). 

The tension in the radial joints will determine the thickness 
of the plates. 

The load on the circumference, or any horizontal joint, 
cut out by a horizontal plane through P P' will be made up of 
— the weight of the cylinder of water whose base is P P', or 



w 



— ^'^ 



a' sm' e {h — a cos ^); the weight of the cone P D P' of 



water, or w . -d^ sin^ 6 . ^a cos ^; and the weight of the 
metal cone below P P', zv' .'-d sin 0^ where zu' = weight of a 
square unit of plate. The last item is comparatively insignifi- 

B^ A 1'^ 

* By the calculus, p at vertex P = — = — - . P L = 2 A; P I = x — A; 

A x^ — A^ 



IN— J. X — A = f7 tan ^; tan (li 



26) — — ; 
a 



. A 



tan 6 



, ^ lan3 ^ -f tan ^ ^ ^ ^ ^ tan ^ 

^ + A = ^ — - — ^j-^ — . _j/ — (x — A) sec 6 = a 



tan"'' — I 



cos H 



tan* 6 — I 

f) = (J tan 6 



= P I. 



230 STRUCTURAL MECHANICS. 

cant, but may be computed when the required thickness of 
plates has been found. 
This weight 

W = 7V r.d^ sin^ [h — f^ cos 0) 

must be carried by stress acting along the elements on the 
circumference P P', that is, inclined along P C, at the angle d. 
From a parallelogram of forces whose diagonal is W, it may 
be seen that this total force on a circumference is W sec 0. 
The section over which it is distributed has a circumference 
2~a sin d. The tension per unit of length of a Jiorizontal 
section ox: joint will therefore be 

w . T.d^ sin^ Ui — %a cos ^) , / 7 o 

.-^—- = -hwa tan 0{ h — %a cos 0), 

cos . 2-a sm <? ^ ^ '^ ^ 

to which may be added \w' a sec for the metal. 
This value will be zero at D, and at C will be 

Jw . D C . tan ^ (/^ — § D C . cos 0). 

At C the tension is decomposed into a vertical component 

z; = iw . D C . sin ^(/2 — f D C . cos 0) = \wr{]i — f D G) 

per linear unit, carried by a circular girder which may itself be 
supported on a wall or on posts. The circular girder in the 
latter case acts as a beam with a torsional moment" added. 
The horizontal component at C will be 

p'— iz£/ . D C . tan ^ sin ^(/^ — f D C . cos t)') = ^wr tan d{h — f D G), 

which causes a compression of p'r in the circular girder. As 
the tension in the tank ring at C is iv[h — D G)r per unit of 
length of a vertical joint, this tension serves to balance more 
or less of p'r, as the construction at C may permit. 
The total weight on the circular girder is 

7V-r\h — § D G) + lu'-r . D C + weight of tank. 

As the stresses found on the joints of the cone are prin- 
cipal stresses, there will be no greater ones at any point. The 
stress on any oblique plane can be readily found, if desired, 
by § 190. 



ENVELOPES. 



231 



Example. — A circular tank, 40 ft. diam. and 40 ft. high, has 
a conical bottom for which ^ = 45° and h = 60 ft. Weight of c. 
ft. of water, 62.5 lbs.; length of element of cone 28.3 ft. Ten- 
sion in radial joint at P, half way up, = 62.5 X 14.14(60 — 
0.707 X 14.14) = 44,187 lbs. per linear ft. = 3,682 lbs. per in. 
of joint. Tension do. at C = 62.5 X 28.3(60 — 0.707 X 28.3) 
= 70,700 lbs. per ft. = 5,892 lbs. per in. of joint. 

Tension in horizontal joint at P = ^ X 62.5 X 14.14(60 — 
f X 14. 14 X 0.707) =3 23,567 lbs. per ft. = 1,964 lbs. per in. of 
joint. Tension do. at C = ^ X 62.5 x 28.3(60 — |- x 28.3 x 
0.707) = 41,242 lbs. per ft. = 3,437 lbs. per in. of joint. v = 
\ X 62.5 X 20(60 — f X 20) = 29,167 lbs. per ft. of girder. As 
the horizontal component equals the vertical component, p'r = 
29,167 X 20 ^ 583,340 lbs. compression in the circular girder. 
Tension in lowest vertical ring of tank = 62.5 X 40 X 20 = 
50,000 lbs. per linear ft. 



Jf a cylindrical 
Fis93. 




240. Tank With Spherical Bottom. 

water tank, of radius r, has a seg- 
mental spherical bottom, of radius r\ 
subtending a central angle of 2 a, 
Fig. 93, the versed sine D E will be 
r\\ — cos a). Sin y = r ^ r' . 
Let the depth of water at the centre 
D be h. 

The tension per iinit of length 
on any radial or meridian joint at 
any point P, the radius to which makes an angle with the 
vertical, will be, by § 229, 

\w\_h — ;-'(! — cos Oj\r' , 

which become \zvhr' at the bottom D, and at C is 

i^[/^ _ r'(i _ cos a)] ;-' = \w{^h — D E)r'. 

A horizontal joint through P must support the weight of the 
cylinder of water of base P P', or 

zv-r'^ sin' 0\Ji _. /( i — cos 0)\ 

the weight of the water in the segment P D P', 

cos 0)\r' — \r\\ — cos 0)1 
\i —cosOy{2 + cos 0), 

and the weight of metal in P D P'. If the weight of a square 
unit of plate is zu', this last quantity is 2za'-r''\i — cos 0)^ 



zu-r'\ I 



= Izv-r" 



232 STRUCTURAL MECHANICS. 

and is comparatively insignificant. Thie weight of water in 
P D P' may often be disregarded also. 

The above vertical forces must be multiplied by cosec to 
give the force exerted, in the direction of a tangent, on the 
circumference P P', and be divided by 2-r' sin to give the 
tension pei' unit of length of a Jiorizontal joint at P, or 

I \ COS (9V'^ 

\wr'\h —;''(! —cos ^)1 -f Uvr""^ . ^ (2 + cos 0\ 



1 



\wr'\]i — ;-'(! — cos 0) -f- \r' tan^ J^ ( 2 + cos 6*)]. 

At C substitute a for 0. 

The vertical component, or load, at C, per unit of cir- 
cumference on the circular girder will be found by dividing the 
weights given above by 2r.r or 2-r' sin a. 

V — \wr\Ji — r'{\ — cos oc)] -\-\ wr'^ ta.n J oc(i — cos a)(2-[-cos <x), 

to which should be added the weight of the tank and of the 
bottom per unit of circumference, the latter being w'r' tan J a. 
The horizontal component /', which causes a compression 
of /'r in the circular girder, will be 

J)' = e; cot a -|- w'r' tan -J a cot a. 

The stresses in the spherical bottom are smaller than 
those in the conical bottom. 

Example. — A circular tank, 40 ft. diam. and 40 ft. high, has 
a spherical bottom for which a = 45°. Then h = 48.3 ft., r' = 
28.3 ft. Weight of a c. ft. of water, 62.5 lbs. Tension in radial 
joint at bottom = Jx 62.5 X 28.3 X 48.3 = 42,645 lbs. per linear 
ft. == 3,554 lbs. per in. of joint. Tension in radial joint at P, 
half way up, where = 22^°, is 40,770 lbs. per ft., or 3,397 lbs. 
per in. of joint. At C, tension is 35,350 lbs. per ft., or 2,946 lbs. 
per in. of radial joint. Tension in horizontal joint at P is 41,760 
lbs. per ft., or 3,480 lbs. per in., and at C is 39,220 lbs. per ft., 
or 3,268 lbs. per in. of joint.- Compression in circular girder 
= 554,700 lbs. . . 

241. Conical Piston. — If the cone E D C of Fig. 94 
represents a conical piston of radius r, subtending an angle 
2 <9, with a uniform normal steam pressure p, per unit of area, 
applied over its exterior or interior, and if the supporting 
force is supplied by the piston rod at D, the compression or 



ENVELOPES. 



233 



tension exerted at P, on a radial section, per unit of length of 
D C, will be, by § 239, if D P = ^, Fig. 92, 

p^ = p p = p . I V — j)a tan d\ 

and the unit stress will be found by dividing by the thickness 
t at P. The maximum unit stress is at C, and will be 



2V' 



tan 



p?^ 



sec 6. 



t sm 6 t 

The vertical force on any horizontal or right section will 
be j9- (r- — a^ sin- 6), which becomes at the vertex -/--p, the 
force on the piston rod. This 
force will be compression on the 
rod and tension in the cone, if p 
acts on the exterior of the cone, 
and the reverse if p acts within 
the cone. The unit stress in the 
metal of the cone at this section will be found by multiply- 
ing this force by sec ^, and then dividing by the cross-section, 
2- a sin^ . /, giving 




'"^^^^ F(g> 94 



i>2 



p r- — a- sm^ d p 



.2 at 



cos 6 sin i) 



at 



r'^ — a- sin^ d 



sm 2( 



which is a maximum at the piston rod. If the radius of the 
rod is r\ a then is r' -=- sin ^, and the stress near the rod is 



_4- sec ^ (r ' — r'^). 

2 7' t ^ 

As the stresses on a horizontal section and a radial sec- 
tion through any point will be of contrary signs at any given 
instant, there will be shearing planes at an angle with p., , by 
§ 192, whose tangent is 1/(^2 -^ Ih)^ '-^nd the value of that 
shear will be y {j^iP^}- 

Example. — Conical piston, Fig. 94, ?- = 24 in., ;-' — 3 in, 



= 69°. Thickness, for ;-' 



in. 



is i^ in.; and, for r' = 



in, is 1.9 in. Steam pressure, maximum difference on two sides, 100 



lbs. per sq. in. For ;-' = 17 in., p^ = 



100 



X 2.79 = 3162 



lbs. per sq. in. ; p.2 



100 



X 2.79 (24- 



^f) 



1570 lbs.; 



2.17.3 

the shear = -1/(3162 . 1570) = 2228 lbs. per sq. in. on a plane 

For r' = 8 in., p, = 1 175 



making 35° 16' with a radial element 



234 STRUCTURAL MECHANICS. 

lbs.; p., — 470 lbs.; and the shear — 743 lbs., at an angle of 
32° 18'. For alternating stresses on steel castings, these values 
are satisfactory. See Example, § 207. 

242. Dome. — A dome, subjected to vertical forces sym- 
metrically placed around its axis, such as its own weight, may 
be treated as follows: — 

Let Fig. 93 be inverted, and let the curve D P C be any 
meridian of the dome. If a horizontal plane P P' is passed 
through the dome, and all the weight from the crown to that 
section is denoted by W, the entire force on the circumfer- 
ential section will be W -^ sin 0^ if is the angle which the 
tangent at P makes with the horizontal. This force, divided 
by the circumference cut out by the horizontal plane, will be 
the compression per linear unit of the circumference, or, if 
divided by the number of ribs in a skeleton dome, will be the 
thrust in one rib. If the horizontal plane is again passed 
through a second point a little nearer C, W will increase, the 
force on the circumferential section will change, since d 
changes, and, as the circumference increases, the compression 
per unit of circumference may be greater, constant, or less, 
depending on the relative changes in the three factors. 

The horizontal component of the entire force in the 
direction of the tangent at P will be H = W cot 0. At the sec- 
tion nearer C, H may be still the same numerically. If so, 
the horizontal band between those two points is in equi- 
librium, having no tendency to move out or in, and hence 
having no hoop tension or compression. If, however, H 
changes, and the change J H is a decrease, a force acting 
inwards must be supplied by that band or hoop, which is then 
in tension. If J H is divided by the circumference, the quo- 
tient will be the normal stress per unit of circumference sup- 
plied by the ring, and the product of this normal stress and 
the horizontal radius of the ring will be the tension in the 
latter, or J H -=- 2 -. The stress per linear unit of a meridian 
section will be found by dividing this expression by the dis- 
tance between the two horizontal planes measured on the 
meridian arc. If, on the contrary, J H is an increase, the 
force to be supplied by the band or hoop acts outwards, and 
compels the hoop to resist compression. 



ENVELOPES. 235 

There is usually a decreasing compression in successive 
hoops from D to a certain section, which for a spherical dome 
of uniform weight is where the tangent has an inclination of 
52° to the horizon, and then increasing tension to C. If the 
dome does not rise vertically at C, and has vertical reactions, 
a strong hoop in tension must be supplied at C. At D a cir- 
cular opening or eye is often made for the admission of light, 
and this opening may be surmounted by a lantern. A strong 
ring at the eye is needed to resist compression. The weight 
of the lantern is easily included with W. 

A ribbed dome may be readily treated in this wa}^ and 
one lune alone considered. The several hprizontal planes 
will then coincide with the purlin rings, and the rib thrusts 
will be taken as parallel to the chords of the successive seg- 
ments of the ribs. In this case, as well as in the preceding" 
one, diagrarns will be more convenient than calculations, and 
sufficiently accurate. 

If the wind pressure on one side of the dome is likely to 
be severe enough to be considered, the trapezoidal panels be- 
tween the ribs and the purlin rings must contain diagonal 
bracing. As all the horizontal component of the wind pres- 
sure above any horizontal plane must be carried past that 
plane, as a shear, to the supporting wall, the rule may be fol- 
lowed that the maximum shear on a thin circular section is 
twice the mean shear. Therefore, put diagonal ties in every 
panel at the level P P', large enough to carry a force whose 
horizontal component is equal to the quotient of the horizontal 
component of all the wind pressure from D to P divided by 
one-half the number of ribs in the dome. As the wind may 
blow from any quarter, both diagonals will be necessary. 

243. Resistance of Thin Ring to a Single Load. — 
The resistance of sewer pipes and similar hollow cylinders to- 
a single load may be found by the following analysis, the 
results being applicable to working loads on ductile materials, 
like steel, and being reasonably correct for breaking loads on 
such brittle materials as cast-iron and vitrified clay pipes. As 
the pipes are comparatively thin, as they are often not very 
homogeneous, and as they vary somewhat from the true cir- 
cular form, it has not been thought necessary to use the exact 



236 



STRUCTURAL MECHANICS. 



value for the moment of inertia of a hollow C3'linder nor to 
take account of the socket. If the section under trial is 
moderately long, the socket will have little or no influence on 
the breaking load. 

If a circular pipe were supported at two points of its 
length and loaded at the middle of the span, the usual for- 
mula for the resisting moment, /I -4- j/j, which would be 
equated with ^W/, might be written, if the cylinder is con- 
sidered to be thin, of a mean radius r and thickness t, since 
jKi = r, and I = -r^t by § 99, VI., 



M 



fr.rt, or/ = 



W/ 



W/ 



\-rH 4S0 1^ 



where So = area of circle of radius r. 

If, on the other hand, the cylinder rests on the bottom 
element and is loaded along the top element, it will fail, if 

brittle, by breaking into four pieces, 
the lines of fracture running approxi- 
mately through the top, the bottom, 
and the extremities of the horizontal 
diameter. If one will press with his 
hand on the top of a moderately 
flexible hoop which rests upright on 
the ground, he will appreciate the 
action of a ring or cylinder under 
two directly opposed equal forces. 
To determine the points of zero 
bending moment or contraflexure, 
B, D, F and G, Fig. 95, or to locate 
what then represents the equilibrium polygon, or action line of 
the forces, it is sufficient to note that the sum of the successive 
changes of inclination, for one quadrant, from A to C, caused by 
the bending moments at successive points, must equal zero, as 
the tangents to the curve at these two points must be unchanged 
in direction. As each change of inclination is directly pro- 
portional to the bending moment at the point, the summation 
of the horizontal ordinates between A and C must be zero. 
As applied in Graphics, Part III., Arches, let a — Q. I, the 
horizontal distance from the ring to the equilibrium line at the 




ENVELOPES. 



237 



extremity C of the horizontal diameter. Let = the angle 
included between O C and the radius to any point N. Then 
will N R, the arm of the constant force JW in B D, be 
r(i — cos 0) — a, which will be of opposite signs for points 
on either side of B. If this expression is multiplied by 7'dO^ 
the length of an infinitesimal arc, and integrated from C to A, 
its integral must be zero. 



[r(i — cos 6*) — a] dO = o = [r{0 

0-57 



Itt,^ = (1- — i)r, or a 



1-57 



sin 0) 
r = o.3634r. 



0-] 



M at side = — ^W X 0.3634;- = — o. 182W/'. 
M at crown = ^W(i — 0.3 634 )r = o.3i8Wr. 

The angle from O C to the points of contraflexure will be 
50° 28', since its cosine is 0.6366. The ratio of the two max. 
bending moments is 7 to 4, or i| to i, the greatest bending 
moment being at A and E. 

244. Resulting Stresses. — Since the resisting moment 
of a rectangular section of width / and depth t is iflf, the 
max. unit tension or compression at the crown will be 



_ o.3i8\\V 
/t — ~ 



if 



6 Wr 

— = 1. 01 ——. 
^ It' 



The max. unit compression at the side, on the inside,, 
will be 



_ iw o. 182W;- . 6 A\' 



= " (* + ■■«. 9 



The max. unit, tension at the side, on the outside, will 
be, if the following expression is positive. 

With ordinary ratios of thickness to radius, the unit stress- 
is much greater at the top and bottom than at the sides, and 
cracking of the clay or cast-iron pipe may first be expected 
at the former points; but failure will probably immediately 
follow at the sides. For a ratio of thickness to radius — ^,. 



2^8 STRUCTURAL MECHANICS. 

the three values become 68.76, 42.24, and 36.24 W -4- /r, 
respectively. 

If these values are compared with the stress from the 
resisting moment for cross-breaking or beam action, it will be 
seen that the stress at A last deduced exceeds the cross- 
breaking stress. To make the two stresses equal it is neces- 
sary that 

W/ Wr ,, 24 r' 



4 Tcr' t It" 

As / in a cross-breaking test of sewer pipe will not exceed 
twenty inches, r must be about 2j t to satisfy this equation. 
The fact is thus made clear that sewer pipes, resting on two 
supports and loaded at mid-span, break in four longitudinal 
pieces as previously described, not as beams, but as cylinders 
under two directly opposed forces, and that tests for cross- 
breaking are not such in fact. 

245. Tests of Clay Pipes. — The following results of tests 
will show the breaking strength as deduced by the formula. All 
the pipes are from the same maker. 



Interior 
Diameters. 


Thickness. 


Weight. 


I 

W 


breaking 
eight, lbs. 


/• 


i8i 


i8i 


Ii^6 


in. 


1 82 lbs. 




4,100 


1,975 lbs. 


i8i 


i8tV 


ife 




178^ 




3' 900 


1,880 


20tV 


20 r\ 


li 




224i 




4,750 


1,960 


24 


23I 


If 




303 




4'775 


i»9&5 


2-li 


23I 


It 




304 




5,200 


2,154 



The mean diameters and a length of 24 in. were used in calcu- 
lating/. Pieces of the 20 in. and 24 in. pipes, subsequently broken 
on two supports with a central load, gave/=: 1,590 and 1,840 
respectively. A piece from the 20 in. pipe, 5 in. long, section 
3 X i^- in- J crushed at 16,000 lbs. per sq. in. 

246. Ring under Any Forces. — If four equal forces were 
applied to the ring of Fig. 95 at points 90° apart, the stresses at A 
and C on the outside would be added algebraically, and similarly 
on the inside circumference. As the moments at A and C are of 
opposite kinds, the new moment would be but three-fourths of that 
now at C, and about three-sevenths of that now at A. A number 
of pairs of equal or unequal forces, of the same or opposite kinds, 
can be readily treated. 

From the action of lateral pressure in diminishing the bending 
moments due to an incumbent load, the ability of a brittle sewer 
pipe to resist a heavy weight of earth, if good lateral support is 



ENVELOPES. 239 

supplied, is made clear. If the pipe is laid in a clay trench, and 
sand or gravel is not packed around it, the larger sizes are liable to 
fracture as above described. As, however, the broken pipes can- 
not spread to any great extent laterally, they may still be ser- 
viceable. 

The unit earth pressure on different planes, for a given ratio 
of /i to/25 maybe found by § 190. By a similar treatment to that 
of § 135, the equilibrium curve for a thin ring under earth pressure 
can then be drawn, and the stress/" caused by a given depth of 
earth can be found. The equilibrium curve, for a pipe buried a 
considerable distance below the surface of the ground, is elliptical. 

This investigation can be extended to rings of other given 
forms, such as the links of chains, either open or studded. In 
chain links the form is rather indefinite, the change of form under 
tension decreases/, and the pull is applied to a, considerable por- 
tion of the curve, and not at a point. The polygon for a studded 
link will be a lozenge or diamond. 

The ultimate strength of chain is stated to be two-thirds of 
that of the original bar, and the safe stress for iron is about 12,000 
lbs. per square inch of both sides for studded links, and two-thirds 
as much, or 8,000 lbs. per square inch, for open links. As links 
differ much in form, only roughly approximate statements can be 
made. 

Examples. — i. A butt-jointed flue in a boiler is 12 in. diam. 
and 14 ft. long. How thick should it be, if 80 lbs. max. steam 
pressure is one-sixth of the collapsing pressure? i\ in. -|-. 

2. What is the net thickness required for a boiler shell 60 in. 
in diameter to carry 120 lbs. steam pressure? What the gross 
thijckness allowing for riveting, and the size and pitch of rivets? 

f in. ; 1%- in. ; f in. : 3 rows, 3 in. pitch. 

3. What weight applied at top of circumference and resisted 
at bottom ought a cast iron pipe, 12 in. diam., \ in. thick, and 
6 ft. long, to safely carry, if / = 12,000 lbs.? 

4. Cast-iron, hydraulic cylinder, 3 in. bore, to raise 15 tons. 
How thick should it be, if unit tension is 7,000 lbs.? 



CHAPTER XIV. 

PLATE GIRDERS. 

247. I Beam. — A rolled beam of I section may be con- 
sidered composed approximately of three rectangles, — two 
flanges, each of area A,, and a web of area A^. The depth 
between centres of stress of the flange sections may be denot- 
ed by /i\ which is also very nearly the depth of the web. 
Then the resisting moment of the two flanges will be /Aj /i\ 
and that of the web, since M for a rectangle is \fbh^, is 
f . JAa/^'. The value for the entire section will be 

M =/(A, -l-JA,)/^'. 

Hence comes the rule that one-sixth of the web may be added 
to one flange area in computing the resisting moment of an I 
beam. The extreme depth of the beam ought not, however, 
to be used for h' . The approximate distance between cen- 
tres of gravity of the flanges will answer, since it is a little 
short of the true value for the flanges and a little longer than 
is correct for the web. 

248. Plate Girder. — A portion of a plate girder and a 
section of the same is shown in Fig. 96. Such a structure 
acts as a beam and is designed to resist the maximum bending 
moments and shears to which it may be liable. It may be 
loaded on top, or through transverse beams connected to its 
web. It is used when the ordinary sizes of I beams are not 
strong enough to resist the maximum bending moment. As 
the flanges may be varied in section by the use of plates where 
needed, as shown at the right, there may be more economy of 
material in using a built beam rather than a rolled one, if the 
required maximum section is large. 

The web A is made of sufficient section to resist the maxi- 
mum shear, and the rest of the material is thrown into the 
flanges B, where it will be farthest removed from the neutral 
axis and hence most eflicient in resisting bending moment. 



PLATE GIRDERS. 



241 



As the thickness of the web plate is usually restricted to one- 
fourth inch, and in girders of any magnitude to three-eighths 
inch, as a minimum, it appears that the material in the web 
practically increases with the depth of the girder. As the 
stress in either flange multiplied by the distance between the 
centres of gravity of flanges resists the bending moment, the 
material in the flanges decreases as the depth increases: — 
hence the most economical depth is that which makes the 
total material in the web as near as may be equal to that in 
the two flanges. Depth of beam contributes greatly to stiff- 
ness, when a small deflection is particularly desirable, and the 
depth may in such a case be so great as to make the web the 
heavier. 

249. Web to Resist Shear Only. — Some engineers 
apply the rule of § 247 to a plate girder, that is, add one-sixth 









c 




c 
: 


: c c 













cocc:cocccccccc"cccccc 


— 


Q 












:>^^-c-crc-c- \ 


D 


c 


A 




-,-.-> - 


-V coooccc -c : :•- : : : : cc oo 


c c.c.c.:_:.f^c^ | 



F=. 



Fig. 96. 



K 



B 



of the web section to the flange section for the resisting 
moment; but the more commonly received practice is to con- 
sider the flanges alone as resisting the bending moment at any 
section and the web as carrying all the shear, uniformly dis- 
tributed over its cross-section, as shown in § %j . As the 
riveted connection between the web and the flanges is not so 
good as solid metal, and as both legs of the angles are inclu- 
ded in the cross-section of the flange, with a moment-arm 
greater than the vertical leg should have, it is advisable not to 
consider any portion of the web as effective for resisting the 

bending moments. 

17 



242 STRUCTURAL MECHANICS. 

250. Compression Flange. — The compression flange 
must be wide enough not to bend sideways Hke a strut be- 
tween points at which it is stayed laterally. As the web 
checks such lateral flexure in some degree, any column for- 
mula to be applied to the flange ma}^ be modified to corres- 
pond. In some cases the value of a in the denominator is 
decreased some forty per cent. One authority gives for the 
allowed unit stress in the compression flange of a railway gir- 
der, consistent with safety from lateral deflection, when the 
flange breadth is b in an unstayed length /, 7, 500 — 

(i 4- ) for wrought iron, with a substitution in the nu- 

merator of 8,600 for soft steel, and 9,400 for medium steel. 
For rolled beams, 8,000, 9,200 and 10,000 are specified; and 
for highway bridges the above values may be increased 25 
per cent. 

The question is complicated by the fact that the unit 
stress varies in the flange plate from section to section unless 
the depth of the girder is made variable to correspond with 
the change in bending moment. A mean value for the flange 
stress may be used. 

One authority specifies that the compression flanges of 
beams and girders shall be stayed against transverse crippling 
when their length is more than thirty times their width. 
Another prescribes that the unsupported length of compressed 
flange shall not exceed twelve times its width; and a third 
specifies the ratio sixteen. Angle irons may be riveted to the 
edges of the flange to stiffen it, or a channel iron with flanges 
turned in may be used. 

251 . Flange Angles. — If the maximum bending moments 
are computed or obtained by a diagram for a number of points 
in the span of the girder, they can be divided by the allowable 
unit stress and the effective depth. The respective quotients 
will be the necessary net sections of the tension flange at those 
points The flange angles E must be large enough to support 
well the compression plates F^ if such plates are required, and 
to be able to transmit the increments of stress from the web 
to such flange plates. Hence the size of the flange angles 
.should be a considerable portion of the largest net section 



PLATE GIRDERS. 243 

found above. For railway girders and floor-beams, it is some- 
times required that these angles should have a section at least 
equal to one-half of the whole flange section required, or be 
made of the largest angles. 

252. Length of Plates. — Inspection of the necessary 
sections will now show how far from the two ends of the 
girder, as at I K, the flange angles, w^ith rivet holes deducted, 
will suffice for the required flange section. From K to the 
corresponding distance from the other abutment the first plate 
must extend, seen on the right in the figure. A reasonable 
thickness being used for that plate, w^ith a deduction for rivet 
holes in the tension flange, it can again be seen where a second 
plate will be needed, if at all. This determination can be 
neatly made on a diagram of maximum moments. Extend 
the plate either way a small additional distance, to relieve the 
angles and assure the distribution of stress to the plate. The 
thicker plate, if there is any difference in thickness, should be 
placed next to the angles. 

The compression flange is usually made of the same gross 
section as the tension flange. The deduction for rivet holes 
in the latter, which is not necessary in the former, compen- 
sates for the slightly lower unit stress allowed for compression. 
If the girder is so long that the plates or angles must be 
spliced, additional cross-section must be supplied by covers at 
the splices, with lengths permitting sufficient rivets to transmit 
the force. Even compression joints, though carefully dressed 
and butted together, are spliced, in good practice. The net 
area of the cover plate and splice angles should be equal to 
that of the largest piece spliced. Only one piece should be 
cut at anyone section, and enough lap should be given for the 
use of sufficient rivets to carry the stress the piece would have 
carried if uncut. 

253. Web: Rivets. — The web carries the maximum 
shear at any point, and this shear is uniformly distributed on 
the vertical section, by § 87. The minimum thickness of web 
before referred to, ^ or | inch, will, for all but long, heavily 
loaded girders, be sufficient to carr}^ this shear. The unit 
shear, if specified, runs from 4,000 to 8,000 lbs. per square 
inch, depending upon the rapidity of imposition of load. As 



244 STRUCTURAL MECHANICS. 

the equal shears on a vertical and horizontal plane in the web 
at any point are equivalent to a pull and a thrust of the same 
unit value at 45° to the horizon, § 189, and Fig. 71, and as 
these inclined stresses must cause horizontal increments of 
stress in either chord, one may conceive the web to be divided 
into square panels and supply enough rivets between the web 
and the angles, for that panel distance and of uniform pitch, 
in double shear or in bearing, to transmit a horizontal force 
equal to the inaximttm shear in the middle of that panel; or 
the maximum shear at any point of the span may be divided 
by the depth of the web and the quotient may be considered 
as the force per unit of length of the flange to be transmitted, 
from which force the pitch of the rivets may be found. As 
the rivets through angles and flange plate are in single shear 
and are in two rows spaced intermediate between those in the 
vertical legs of the angles, the same pitch will be correct, 
unless the flange plates are deficient in bearing area. The 
maximum shear at any point in the span will occur when the 
longer segment is loaded, if possible. 

Make the pitch of rivets in inches and eighths, not deci- 
mals; do not vary the pitch frequently, and do not exceed a 
six inch pitch, so that the parts may be kept in contact. If 
flange plates are wide, and two or more are superimposed, 
another row of rivets on each side, with long pitch, may be 
required, to ensure contact at edges. Care must be taken 
that a local heavy load at any point on the flange does not 
bring more shear or bearing stress on rivets in the vertical 
legs of the flange angles than allowed in combination with the 
existing stress from the web at that place and time. 

Webs are occasionally doubled, making box girders, suit- 
able for extremely heavy loads. The interior, if not then 
accessible for painting, should be thoroughly coated before 
assembling. 

If the web must be spliced, use strips for that purpose, 
having the proper thickness for rivet bearing and enough rivets 
to carry all the shear at that section; do not make a splice 
with a T iron. 

254. Stiffeners. — At points where a heavy load is con- 
centrated on the girder, stiffeners C, consisting of an angle 



PLATE GIRDERS. 245 

iron on each side, should be riveted, to prevent crushing of 
the web under the local load and to distribute such load to 
both flanges. They should for a similar reason be used at 
both points of support D. These stiffeners maybe computed, 
when necessary, as a + shaped column of four equal arms, 
each of the width of an angle leg. 

Since the thrust at 45° to the horizontal tends to buckle 
the web, and the equal tension at right angles to the thrust 
opposes the buckling, it is conceivable that a deep, thin web, 
while it has more ability to carry such thrust as a column or 
strut than it would have if the tension were not restraining it, 
may still buckle under the compressive stress; and it is a ques- 
tion whether stiffeners may not be needed to counteract such 
tendency. They might be placed in the line of thrust, sloping 
up at 45° from either abutment, but such an arrangement is 
never used. They are placed vertically, as at C, and spaced 
bv a more or less arbitrarv rule. 

A common formula is: — The web of the girder must be 
stiffened if the shear per square inch exceeds 

r d^ \ 

12.000 ^ I I H :, I . 

V ' 3,000/V ' 

where d — clear distance between flange angles, or betweeji 
stiffeners if needed, and / = thickness of web. Another rule 
calls for stiffeners at distances apart not greater than the 
depth of the girder, when the thickness of the web is less than 
one-sixtieth of the unsupported distance between flange angles. 
The above formula may be written in terms of the slanting 
distance, the real strut length, but is more convenient as it 
stands. Experience appears to show that stiffeners are not 
needed at such frequent intervals as the formula would 
demand. An insufficient allowance for the action of tension 
in the web in keeping the compression from buckling it, is 
probably the cause of the disagreement. 

Stiffeners may be offset at the ends, as shown in the 
figure at E, or filling pieces may be used under the angle 
stiffeners to avoid the offset; in all cases they should be 
tightly fitted between the two flanges. 



246 STRUCTURAL MECHANICS. 

Sometimes the depth of the girder is varied to approxi- 
mate to the elevation of a beam of uniform strength. 
For unit stresses; see § § 17 1-7. 

Example. — A plate girder of 30 ft. span, load 3,000 lbs. per 
ft, / = 35,000 lbs. per sq. in. W =: 90,000 lbs., and M max. = 
i-W/ =: 4,050,000 in. lbs. Assume extreme depth as 42 in., effect- 
ive depth, 39 in. Net flange section at middle = 4,050,000 
-f- (39 . 15,000) = 7 sq. in. A f in. web, 42 in. deep, will have 
15 J sq. in. area. Two flanges, each 7 sq. in. net -j- allowance for 
rivet holes, will fairly equal the web. Use f in. rivets. 

Let the flange angles be 2 — 4 x 3 X f in. = 4.96 sq. in. 
Deduct 2 holes, |^ x g^ = 0.66. Net plate = 7 — 4.3 = 2.7 sq. 
in. A plate 9 x f = 3.f sq. in.; deduct two holes = 0.66, leaving 
2.71 sq. in. Two angles and plate, gross section = 4.96 -f- 3.37 
— 8,33 sq. in. Resisting moment of net section of angles ^=. 
4.3 X 15,000 X 39 ^ 2,515,500 in. lbs. Such a bending moment 
will be found at a distance x from either end, given by P^ x — 
\ . 3,000 x^ = 2,5 15,500. X -— 5.9 ft. . • . Cut off the plate 5 ft. 
from each end. 

Shearing value of one f in. rivet at 10,000 lbs. per sq. in. 
= •4,400 lbs. Bearing value in f in. plate, at 20,000 lbs. = 5,6co 
lbs. Max. shear in web = 45,000 lbs. Pitch for flange angles, 
since bearing resistance is less than double shear, = 5,600 . 39 
-^ 45,000 = 4.85 = 4J in. Make 3 in. pitch for 2 ft., then 4f in. 
for 6 ft., then 6 in. pitch to middle. Rivets in end web stiffeners, 
45,000 -^ 5,600 =r 9. Max. shear in web := 45,000 -^ (42 . -|) 

= 2,780; 12,000 -f- I I -| 2 I = 2,950, since </ = 42 — 6. 

No other stiffeners needed. By the other rule 36 -7- f = 96, and 
stiffeners are needed. 



CHAPTER XV. 

EARTH pressure: RETAINING WALL! SPRINGS: PLATES. 

255. Pressure of Earth. — The stabihty of a mass of 
earth and the resistance that must be offered by a retaining 
wall to the thrust of a bank can be determined by the princi- 
ples of Chap. XL, if it is assumed that the .particles of earth 
are held in place by friction alone. The adhesion arising 
from the presence of a little moisture is neglected, as always 
uncertain in amount and sometimes possibly absent. Such 
adhesion would diminish the- pressure against the wall. If the 
earth is saturated with water, so as to be reduced to mud, it 
will press normally against the. wall as does a fluid, and with 
a pressure which is to that of water as the weight of a cubic 
foot of mud is to one of water. If friction alone operates to 
keep the particles at rest, the greatest possible obliquity of 
pressure from the normal, consistent with equilibrium, on any 
plane in the mass of earth, cannot exceed what is known as 
the angle of repose ; for, if it did, sliding would take place 
along that plane. 

Let a plane be passed through P, Fig. 97, parallel to the 
surface of the ground D K. The pressure on every square 
foot of this plane is vertical, and due to the earth above it, of 
depth K P. But the prism of earth resting on a square foot 
of this plane has a smaller horizontal section than one square 
foot, and the ratio of the unit vertical pressure, on the plane 
through P, to the weight of a vertical column of earth one 
square foot in cross-section will be that of the normal P E, 
drawn from P to D K, to P K. Hence, revolve P E to P G, 
and G P will represent, in feet of earth, the pressure per 
square foot of the plane through P parallel to the surface of 
the ground. 

If the principal stresses p^ and p^ were known, P M could 
now be laid off on the normal = i(/i -]- p^^ and M G would 



248 



STRUCTURAL MECHANICS. 



be l{pi — pi) to close on G P or /, § 190. But, as stated in 
a preceding paragraph, the greatest obhquity of stress from 
the normal to any plane cannot exceed the angle of repose of 
the earth. Hence, if P I is drawn, making that angle with 
P N, the distance M G must, if applied at M O, make MOP 
a right angle. Therefore, find by trial a centre M on N P 
from which a semicircle N G O can be drawn through G and 
tangent to the line P I. The point M can readily be located 

very closely. P M will 
then be l{p^ -f p^) and 
M G, i(/i — /J. By 
§ 193, the direction of 
pi will be parallel to the 
line M L, drawn bisect- 
ing the angle N M G. 
It may be noted 
that the two principal 
stresses act on the right- 
angled faces of a small 
triangular prism at P, 
the other face being 
F B! ^ parallel to the surface 
of the ground; that p^ 
is fhe least possible 
pressure which is con- 
sistent with equilibrium, 
and that it is the one 
Y exerted by earth at rest 
under the action of its own weight only. Blows applied to 
the surface of the ground, the vibration set up by railway 
trains, and similar causes will probably increase the pressure 
and should be allowed for as is a live load on a bridge. 

256. Pressure Against a Wall. — To find the centre of 
pressure and maximum unit pressure with its direction on any 
bed-joint of a retaining wall, weighing :£/' per cubic foot, press- 
ed at the back by earth of a weight zu per cubic foot and of a 
given angle of repose, proceed as follows: 

The bed-joint is A B, Fig. 97, carrying the weight of 
masonry A B C D, whose centre of gravity is at T. To find 




EARTH PRESSURE: RETAINING WALL. 249 

T, draw diagonals A C and B D; bisect each at i and 2 re- 
spectively; lay off A-4 = C-5 and B-3 = D-5; connect i with 
4 and 2 with 3; these connecting lines will intersect at the 
centre of gravity, T. At a point P, the same distance K P 
below the surface of the ground that A is, make the con- 
struction just described in § 255, and bisect the angle N M G 
by L M. 

Find the unit pressure and its direction at A on the plane 
A D, by § 190, as follows: — Draw A Q perpendicular to A D; 
lay off A Q = P M = i(/i + /s); draw A S parallel to L M, 
that being the direction of p^; from Q as centre, with radius 
A Q, draw an arc cutting A S at S; dravv' Q S and lay off on 
it from Q, M G = Q R = J(/i — p^^\ connect R with A, and 
R A will be the direction of the pressure at A on the back of 
the wall, and its magnitude per square foot in terms of cubic 
feet of earth, so that, if R A is measured by the scale of the 
drawing and multiplied by the weight of a cubic foot of earth, 
the pressure on the back of the wall per square foot at A will 
be given. 

As the pressure at the back increases regularly with the 
distance below the surface of the ground, the centre of pres- 
sure will be at X, one-third of the slant height from A, and 
the total earth pressure against one foot in length of the wall 
will be i(A DxA R). 

257. Resultant Pressure on a Joint. — Draw X U W 
through X, parallel to R A, and let fall T V vertically through 

T; make U V = (A B -}- C D) — , and U W = A R ^. " 

zv \^ r 

Complete the parallelogram U V Y W. U Y will be the 
direction of the resultant pressure on the bed-joint A B, and 
the point Z where it cuts the joint will be the centre of resist- 
ance or pressure. The total pressure on the joint will be 
found by multiplying U Y by one-half the height of the wall 
above A B and by the weight of a cubic foot of earth. 

AD 

■^^The ratio ^ ^ may be called unity without serious error, unless the wall 
C F 

has a strong batter at the back. By the use of the above factors, U V represents 
the weight of the wall and U W the total earth pressure at back of same. 



250 STRUCTURAL MECHANICS. 

258. Maximum Stress on Joint. — If it is thought that 
this centre of pressure is too near the front for safety, or too 
near the middle of the joint for economy of masonry, change 
the section by drawing D A' or D A" and try again. A sec- 
ond trial will usually suffice. If the distance B Z is more 
than one-third of B A the maximum unit pressure per square 
foot, at B, is, by § 138, 



P 



2 . total pressure /' sBZ'A 



A B 



(^ - Fi )■ 



If the distance B Z is less than one-third of B A, the 
maximum unit pressure at B, supposing the cement in the 
joint to offer no resistance to tension, is 

_ 2 total pressure 



P = % 



B Z 



If this pressure is greater than the masonry can safely 
resist, make A B wider and try again. The wall will be sat- 
isfactory to many engineers if B Z is somewhat greater than 
JAB. A margin is thus left for an increase of pressure be- 
yond the least pressure here used. The obliquity of U Y to 
the perpendicular to A B will determine the tendency of the 
wall to slide forward. If such sliding seems likely to occur, 
the bed-joint A B may be inclined backwards. The above 
constructions are simplified when the surface of the ground is 
horizontal, and also when it slopes at its angle of repose . 

259. — Remarks: Special Case.— A little study of the fig- 
ure will show that a slope at the back of considerable amount 
has the advantage of increasing the obliquity of the pressure 
against the wall, and hence of throwing Z nearer the middle 
of the joint. 

A rough rule for a retaining \yall, when the ground sur- 
face is level, is to make the average thickness one-third of 
the height. For a wall 15 ft. high, and 2^ ft. wide at coping, 
this rule would make the base 7j ft. thick. If the earth to 
be supported by a wall rests on an inclined stratum which may 
be penetrated by water to such an extent as to make the in- 
clined plane slippery, the component of the weight of all the 
earth above that plane in a direction tangent to it may be 




SPRINGS. 251 

brought against the wall, and its point of application will be 
in a line drawn through the centre of gravity of that mass, 
parallel to the plane ol sliding. Such pressure may be too 
great for any reasonable wall to resist, when it is forced to 
hold up an entire hillside. Expedients should be resorted to 
in such a case to thoroughly drain the troublesome stratum, 
or to build a bank of stone and coarse gravel at the toe of the 
slope behind the w^all. 

SPRINGS. 

260. Straight Spring. — For a spring of varying section, 
see § III. If a beam of uniform section, fixed at one end, 
has a couple or moment applied to it, 
Fig. 98, in place of a single transverse 
force, it will, as shown in § 103, bend 
to the arc of a circle. The deflection 
will be, if / is the length of the beam, 

V =— ^— . Since the stress in the extreme fibre, 
2E I 

For very small displacements, the work done by the 
rotation of the couple M will be, since M -f- / is the equiva- 
lent force at each end, and dv the small distance through 
which the force moves at any instant, 

.V 1 [^ ^ 4E I /^ 2EI , /I f 

Work = . /- ^. = ^^ /.^. = ^^ .- = _ . -. 



For a rectangular section bh, these quantities become 
^ 6M fP E/i , 

f- f- 

Work = bhl . ^ = Volume . -^. 
6E 6E 

For a circular section the number 6 in the last expression 
will be replaced by 8. 



252 



STRUCTURAL MECHANICS. 



261. Coiled Spring. — In practice the rectangular or 
cylindrical bar is bent into a spiral and subjected to a couple 
M = Fa, Fig. 99, which, as a couple can be rotated in its 

plane without change, acts equally at all 
sections of the spring. The developed 
length of the spiral is /. 

262. Helical Spring. — A cylindrical 
bar whose length is / and diameter d, 
when fixed at one end and subjected to 
•^'S-^-^- a twisting moment T = P<^ at the other, 

if the elastic limit is not exceeded, by § 91, is twisted through 

33T/ 




an angle 



-Cd- 



The work expended in the torsion is 



/ 



TdO 



64/ 



From 



9I: 



2qd_^ 
~ Qd' 

Work 



and therefore 



/. ^ = Volume 
4C 



If C = I E and q^ =4/. work = vol- 



/■2 

nme . ~ — , while 
5 E 



for flexure, as just 



shown, work = volume 



6E' 



a smaller 



1l 

4C' 



quantity, so that the torsional moment 
does more work than the bending mo- 
ment. 

If this bar is bent into a helix and 
the force P is applied at the centre, in 
the direction of the axis of the cylinder, 
Fig. 100, to a horizontal arm whose 
length is a, the arm possessing sufficient 
stiffness to be not appreciably bent, the 
moment Ya will twist the bar throughout its 




length. 



Then 



^1 = 



16 P^ 



-.73 



-d 



or P 



16 a 



SPRINGS. 253 

The deflection of the sprino^ vs^ v = a d, since, as the force P 
descends, the spring descends, and the action is the same as if 
the spring remained in place and the arm revolved through an 
angle 6. The force P is too small to cause any appreciable 
compression (or extension) of the material in the direction of 
its length. 

32 Va^ I lal q^ \-nd^ q^ 

^° ^ ^ ■ cT^ ^ ~d ' 0,"^' ~ir ■ C ' 

if ;/ = number of turns of the helix, and 1=2 -an. 

P may be tension in place of compression. 

If the section of the spring is not circular, substitute the 

proper value of q^ or the resisting moment from § 92. If the 

r d'\ 

rod is hollow, multiply the exterior volume by j i ■ — -^ I . 

For a square section and a given deflection, P will be about 
65% of the load for an equal circular section. C for steel is 
from 10, 500, 000 to 12,000,000. 

Example. — A helical spring, of round steel rod, i in. diame- 
ter, making 8 turns of 3 in. radius, carries 1,000 lbs. 

16.1,000.3.7 _^ 4.22.8.9.15,273 

q. = =1^,273. z; = = 1-15 in. 

22 7- 1 • 12,000,000 

263, Circular Plates. — The analysis of plates supported 
or built in and restrained at their edges, and loaded centrally 
or over the entire surface, is extremely difficult. The follow- 
ing formulas from Grashof's "Theorie der Elasticitat und 
Festigkeit" may be used. The coefficient of lateral contrac- 
tion is taken as \, or in = 4. 

I. Circular plate of radius ;- and thickness /, supported 
around its perimeter and loaded with zv per square inch. 

f^ — unit stress on extreme fibre in the direction of the 
radius, at a distance x from the centre. 

/y = unit stress perpendicular to the radius, in the plane 
of the plate, at the same distance x from the centre. 

A- ^^g ^,K^ ^ o^^ )^ A- J, 8 t'^^ ' :'^- 

117 i-'r' 189 wr' 

A = Amax. (forx = o) = --^^. .-, :. ^^ . 



254 



STRUCTURAL MECHANICS. 



For the same value of /, the max. stress is independent 
of r, provided the total load iv-r'^ is constant. 

II. Same plate, built in or fixed at the perimeter. 

At the centre, f^ — f^. At the circumference /y is zero, and 
f^ is max. 



45 ^^ 
Amax. =^.^ 



45 wr"" 



256 Y.f 

III. Circular plate supported at the perimeter and carry- 
ing a single weight W at the centre. Loaded portion has a 
radius r^. 



/x 



45 "^ n 

~ 72- Gog 



32 



*); /. = f:?(log-^ + f). 



These expressions become maxima for x 
second is the greater. 

117 Wr^ 



and the 



For values oi r -h- r^ 
/max. 



° ~ 64- E/^ • 

10, 20, 30, 40, 50, 60, 

1.4 1.7 1.9 2.0 2.1 2.2 W^/^ 



If r^ — o, the stress becomes infinite, as is to be expected, 
since W will then be concentrated at a point, and the unit 
load becomes infinitely great. It is not well to make r^ very 

small. 

IV. Same plate, built in or fixed at the perimeter. 



/x 



45 ^^ /I ^ \ X 

T (log-— 0; /y 



32 



X 



45 W r 

32- f' ^x 



The maximum value of/is/y, for ;ir = r^ 

45 Wr'^ 



W^ ^ /'-^ 



°~: 647: E/^' 

For values of r -^ r^ — 10, 20, 30, 40, 50, 60, 
/max. = i.o 1.3 1.5 1.6 1.7 i.g 

264. Rectangular Plates. — The problem of the resist- 
ance of rectangular plates is more complex than that of circu- 
lar plates. Grashof gives the following results: 



PLATES. 



255 



V. Rectangular plate of length a, breadth b and thick- 
ness t, a > b, built in or fixed at edges and carr3dng a uni- 
form load of w per square inch. 

b"" . wa^ _ a'' . wb- 

2 (a' + b')f ' ^^ ~~ 2 (a' + b') f ' 

The most severe stress occurs at the centre in the direction b, 
that is, on a section parallel to a. 



A = 



li a 



b, f = 



wa' 



\f- 



The deflection at the centre is z'n = 



a' b' 



w 



a' ^ b' 32 E/ 



3 ' 



and for 



a square plate, 



wa 



64 Y.f ' 

VI. Plate carrying a uniform load of if per square inch 
and supported at rows of points making squares of side a. 
Fire-box sheet with staybolts. 

_ 15 laa^ 



_ 15 wa'^ 



Navier gives formulas for rectangular plates which are sup- 
posed to be ver}^ thin. Approximate values from those form- 
ulas are as follows: 

VIL Rectangular plate, as in V., but supported around 

the edges. 

a^b"^ w _ a^b^ w 

VIII. Rectangular plate, supported at edges and carry- 
ing a single weight W at centre. 



/ 



a^b 



W 



2.2; 



0.46 



a'lr 



W 



i^a' -\- b-y f-' ° ^'^^ {a- -^ b-)-Y.r 

For the same total load, f is independent of the size of 
the plate, provided the ratio a to b and the thickness are 
unchanged. 

Example. — K steel plate, 36 in. square and \ in. thick, sup- 
ported at edges, carries 430 lbs. per sq. ft., or 3 lbs. per sq. in. 
/= 0.92 . 1 . 36 . 36 . 3 . 16 = 14,300 lbs. 

0.19 36^ . 3 . 4^^ _ ^ 



V = 



30,000,000 



^ \\\. 



256 



STRUCTURAL MECHANICS. 



265. Corrugated Iron. — The curves to which corrugated 
sheets are bent can be defined only by their depth and width. 
Since it is necessary to determine the moment of inertia of 
the cross-section of the sheet transversely to the corrugations, 
in order to write an expression for the resisting moment, or to 
determine the stiffness, any curve may be assumed which will 
be a good approximation to the probable one, and for which 
the moment of inertia can be obtained in the above-men- 
tioned terms. The cycloid is such a curve. 

Let A B C, Fig. 10 1, be the curve described by the point 
B of the circle B D, as it rolls on the line ADC. Any arc 
B I of the cycloid will be equal to twice the length B H 
of the chord of the" generating circle drawn from B to the 
point H which B will occupy on the circle, when in rolling it 

has reached I of the 
Fig 101. cycloid. For, as the 
-L \ ^\ circle turns at any in- 

stant about the point 
,F of contact with A D C, 
the point H moves 
along H K. with radius 
D H, for a small displacement, describing I N. But H K is 
equal to twice the amount by which B H' exceeds B H; there- 
fore any arc B I is equal to twice the corresponding chord B H, 

Let B I = i- = 2B H; B F = ^ = 2B G. 

B H2 = B L^ + L H^ and L H^ =r B L . L D. Therefore 
BH- = BL' + BL.LD = BL(BL + LD) = BL.^, 

where d = diameter of generating circle. Similarly 

B G'' = B M . d. But B E . B F = r' = 4B G' = 4B M . 

E I . I F = (c-s)(c^s) = c' — r= 4(B G"— BH^') 
= 4(B M — B L)^/. 




d. 



B M — B L 
BM 






Thus it appears that any portion of a cycloidal arc above 
an arbitrai-y line E F, drawn parallel to A C, may be taken, 
and the ordinate y from that line to any point I has the ratio 



SPRINGS. 



257 



to the ordinate at the middle, \Ji, of the product of the 
respective segments I E and I F to B E and B F, into which 
those ordinates divide the arc E B F. Then, for arc E B, of 
thickness t, 



I 



r.7 i^^^ fr ^'\^, th- r 2S^ , s' ^-^ 2 ,, 



As c^ = cross-section, r^ = — /r. 

The depth of the corrugation is /i, the breadth of the 
sheet in the undulating line is c. The breadth straight across 
may be used by modifying slightly the value of /. 



CHAPTER XVI. 

DETAILS IN WOOD AND IRON. 

266. General Principles. — In designing and executing 
all kinds of joints and fastenings the following general princi- 
ples, as given by Rankine, should be used as a guide: — - 

To cut the joints and arrange the fastenings so as to 
weaken the pieces of timber they connect as little as possible. 

To place each abutting surface in a joint as nearly as pos- 
sible perpendicular to the pressure which it has to transmit. 

To proportion the area of each such surface to the pressure 
which it has to bear, so that the timber may be safe against 
injury under the heaviest load which occurs in practice; and 
to form and fit every pair of such surfaces accurately, in order 
to distribute the stress uniformly. 

To proportion the fastenings, so that they may be of 
equal strength with the pieces which they connect; and to 
place them so that the}^ may not shear out of the timber nor 
crush the fibres. 

The same principles are applicable to metallic construc- 
tion. 
I 267. Framing of Timber: Splices. — Sketches XI. to 

XVI. in Plate II. represent different methods of splicing a 
timber tie. In each case the smallest cross-section of the 
timber determines the amount of tension that can be trans- 
mitted. The shoulders are in compression, and the longitud- 
inal planes between the shoulders are in shear. In XL, for 
equal strength, the depth of the . two opposite shoulders or 
indents should be to the remaining depth of the timber as the 
safe unit tensile stress is to the safe unit compression along 
the grain. The shearing length, on either timber or clamp, 
should be to the depth of shoulder as the safe unit compres- 
sion is to the safe unit shear. In actual practice, unless con- 
siderable dependence is placed upon the resistance of the 



DETAILS IN WOOD AND IRON. 259 

bolts against shearing through the timber, the spHce should be 
much longer than shown. If the two clamps are of stronger 
wood than the main timber, they need not together have so 
much depth as the net depth of the timber. The iron strap 
in XIV. illustrates the same principle. The bolts are usually 
small, and serve mainly to balance the couples set up on each 
clamp by the pressure on the shoulder and the tension in the 
neck. The modification in XII. permits the introduction of 
the bolts without reducing the net section of the timber. In 
XIIL, each indent is only half the previous depth, with 
obvious economy of the main timber, and increase of shear- 
ing area of clamp and timber without lengthening the clamps. 
It is much more difficult to fashion, however, and it is not 
probable that both shoulders on one half will bear equally. 

XV. and XVI. are scarfed joints. The tension sections, 
the compression shoulders and the longitudinal shearing planes 
should again be properly proportioned here. In XV., but 
one-third of the timber is available, if unit tension and com- 
pression have the same numerical value, while in XVI. one- 
half of the stick is useful; but the latter joint is more trouble- 
some to fashion. The bolts serve to resist the couple which 
tends to open the joint, and, by resisting it, cause a fairly uni- 
form distribution of stress in the critical section. The bolt 
holes do not weaken the timber. Sometimes the extreme ends 
of the scarf are undercut to check the tendency to spring out 
when the bolts are not used. Keys may be driven through 
places cut for them at the shoulders. The joint can then be 
readily assembled and forced to place. These sketches show 
that timber, although possessing good tensile strength, is ill- 
adapted for ties, on account of the great loss of section in 
connections and joints. 

268. Struts and Ties. — The connection of a strut and 
tie in wood is illustrated in II., III., IV. and VII. The 
shrinkage of the pieces of II. in seasoning tends to open both 
portions of the joint by changing the angles; but the bearing 
of the strut is still central, if only on a small area. The com- 
pression of the tie across the grain may be large in such a case, 
and the introduction of a block, as in IV., will remedy such a 
difficulty as well as that from shrinkage. The block below is 



2 6o ' STRUCTURAL MECHANICS. 

the wall-plate, for distributing the truss load along the wall. 
It is subjected to compression across the grain. 

If the shearing area to the left in these four cases is not 
sufficient, the bolt or strap is a wise provision to take up the 
horizontal component. The bolt, if a little oblique to the 
strut, as shown, holds at once by tension, to some degree, 
and not alone by shear. It also relieves the smallest section 
of the tie from a part of the tension. The square shoulders of 
III. are good, if the timber is seasoned, as the bearing is then 
over the whole end of the strut, and the tie is not weakened 
any more than in II., while the joint is more simply laid out. 
The strap of VII. gives a satisfactory bearing for the strut, but 
the fastenings of such a strap are often weaker than the strap 
itself. The holes in it may well be enlarged hot, without 
removal of metal and diminution of cross-section. 

In VIII., IX. and X. are shown connections of struts 
which may at some time be called on to resist tension, or 
which may be relieved of stress and become loose. The tenon 
in VIII. rnust be pinned to carry tension; and the pin will 
resist but little before shearing out of the tenon or splitting off 
the side of the other timber by tension across the grain. The 
tenon should be fashioned as indicated, with sufficient area at 
the left hand edge to carry the perpendicular component of 
the thrust of the strut as compression across the grain, and 
sufficient cross-section not to shear off. The size of the strut 
must be determined, not only by the column strength, but by 
the area necessary to prevent crushing the piece against 
which it abuts. This remark applies to IX. and X. also. 
The ability of IX. to carry tension depends on the resistance 
of the nut, which is slipped into a hole at the side, to shear- 
ing out along the strut, or crushing the fibres on which it bears, 
the latter method of failure being the more likely, unless the 
nut is quite near the end of the strut. The strap on X. is 
very effective, and the arrangement, if inverted, will serve as 
a suspending piece, although a rod is better. Many of these 
connections are serviceable in other positions. 

To keep a strut from crushing the side of a timber, a con- 
nection may be employed, as in the lower part of I. This 
device may be economical, if a number of such joints are to 



DETAILS IN WOOD AND IRON. 261 

be made, and it is superior to a mortise in work exposed to 
the weather, as there is no place for water to lodge. The 
post in XVIII. is capped by a similar device for distributing 
and thus reducing the unit pressure on the other piece. 
Lateral displacement is provided against in both cases b}^ ribs 
on the castings. 

Strut connections are shown in XIX. and XX., with a 
tie rod in addition. The broad, flat washer reduces the unit 
compressive stress on the wood under it: the lip keeps water 
out of the joint. Shrinkage and a slight deflection of the 
frame under a load will cause the mitre joint in XIX. to bear 
at the top only, throwing the resultant stress out of the axis 
of the respective compression members, §§ 137, 151, and 
causing the unit compression at top edge of the joint to be 
very high. The joint in XX. gives a better centre pressure, 
and is easily made; the upper piece is simply notched for one- 
half its depth, and the upper and lower edges come on the 
mitre line of XIX. The connection of XXL, by the insertion 
of an iron plate or a block of wood, secures a certain con- 
tinuity or rigidity in the joint, to resist a moderate amount of 
bending moment. The two pieces might have been halved 
together. XXVI. is like VIII. , without provision for tension, 
which is usually unnecessary. The roof purlin with its block 
is also shown in relative position. 

269. Beam Connections, — In I. and XVIII. are shown 
supports of beams on posts. The double or split cap of I. is 
serviceable where several posts are to be connected laterally, 
as in a trestle bent, and it is desired to do away with mortises. 
A mortise and tenon of usual proportions are shown in XVII. 
Bolts should be put transversely through the caps and top of 
the post. A comparatively wide bearing for the beam, with- 
out the use of large timber caps, may be here secured. 
Lateral bracing, as in XXVIII., will be needed. An indirect 
and intermediate support for a beam, by two inclined braces, 
is seen in XXV., and the reverse case is represented in XXVII. 
The ordinary wall bearing for joists may be seen in the lower 
left-hand corner. The slanting end is a wise provision to pre- 
vent harmful action of the loaded joist on the wail, and it 
promotes ventilation of the timber. 



262 STRUCTURAL jMECHANICS. 

The usual way of connecting two floor joists or beams, 
when their upper surfaces are to be at one level, is drawn in 
VI. The nearer the mortises are to the neutral axis, the less 
the weakening of the pieces in which the}/ are cut; on the 
other hand, the farther the two tenons are apart, the more 
firmly is the tenoned joist held against lateral twist. The 
shouldered tenon, indicated by the dotted lines at the left, is 
designed to attain both objects, to weaken the mortised piece 
as little as possible and to have a considerable depth of tenon, 
as well as a long tongue projecting entirely through. The 
work of framing is considerably more than in the former case. 

270. Wooden Built Beams. — If seasoned material is 
at hand, and large timbers are too expensive, a useful beam 
may be built up by placing planks, from two to four inches 
thick, edge to edge, and then thoroughly nailing or spiking 
boards on both sides at an angle of 45° with the length 
of the beam, and sloping in opposite directions on the two 
sides. The planks will carry the direct stress due to the bend- 
ing moments, and the boards will resist tension and compres- 
sion equivalent to the shear, as shown in § 186, By due 
regard to jointing and nailing a beam of considerable span 
may be made at moderate cost. The construction can be 
doubled if necessary. 

Another compound beam is seen in XXV. The keys and 
bolts resist the shear along the neutral axis; the horizontal 
sticks are butted together on the compression side, and are 
strapped by the metal clamp indicated to carry tension, if 
necessary. The small block behind the clamp keeps it in place. 

A timber beam has been fashioned like a plate girder, 
with a close web of diagonal boards spiked at 45°, and flanges 
of planks connected by other planks occupying the place of 
the flange angle irons. Its efficiency is uncertain. The old- 
fashioned plank lattice bridge was very cheap, where lumber 
was plenty, as the cost of construction was very small. 

271. Curved Beams. — -Planks placed side by side, as in 
XXII., cut to the form of a curved beam or arched rib, and 
bolted together to prevent individual lateral yielding, are 
quite effective, if the grain of the wood does not cross the 
cu]'ve too obliquely. Hence, when the curvature is considera- 




"'S' 



XVT. 




264 STRUCTURAL MECHANICS. 

ble, it may be advisable to use short lengths, which must 
break joint in the several parallel pieces. It is well to make 
a deduction of one piece in computing the strength of the 
member at any section. The ratio of strength of this com- 
bination, when well bolted together, to that of a solid stick 
may be considered to be as ;/ — i to n, where n is the number 
of layers. 

If the planks are bent to the curve and laid upon one 
another, as in XXIII., this combination is not nearly so effec- 
tive as the former, but it can be more cheaply made. The 
lack of efficiency arises from the unsatisfactory resistance 
offered to shear between the layers by the bolts or spikes. 
The strength to resist bending moment will be intermediate 
between that of of a solid timber and that of the several planks 
of which it is composed, with a deduction of one for a prob- 
able joint. It may be taken, if the beam is well bolted, as the 
mean of the two values, or as i^n^ -\- 71 — i ) -^ 2n^ of the resist- 
ance of a solid stick, if n = number of layers, or as (;z + i) 
-i- 2«, when one layer is not deducted. 

Example. — An 8 in. by 8 in. beam is made of four 2 in. planks 
on edge, with a joint every 3 ft. Its resisting moment will safely 
be \fbh' , (n—i) ^ n = ifS' . 1 = 64/ 

A similar curved beam is made from eight i in. boards, bent 
to curve and well nailed, one on top of the other. Its resisting mo- 
ment will be \fbh^. {n^ -i- n — i) ^ 211^ — \j Z^ . 71 ^ 128 = 47 J/. 

If the curved member has a direct force acting upon it 
and a moment arising from its curvature, the treatment will 
follow the same lines; but the joints, if there are any, will be 
more detrimental in case there is tension at any section. 
Such curved pieces are sometimes used in open timber trusses 
for effect, but their efficiency is low on account of the large 
moment due to the curvature. XXII. is the stiffer. 

The joints and connecting parts in all timber construction 
should be proportioned in detail for such tension, compression 
and shear as they may have to withstand. Often the three 
kindsof stress occur in different parts of one joint or connection. 

272. Iron Roof Truss.— Joints I. to IV., Plate III., 
represent ways of connecting the several pieces of a compara- 
tively light roof-truss. All the members are made with angles 



DETAILS IX WOOD AND IRON. 265 

and at several points both legs of the tension angles are fast- 
ened. Joint I. comes between II. and III., and IV. comes 
perpendicularly opposite it. The number of rivets in each of 
the ties and centre member of II. depends upon the force in 
the particular piece and the rivet shearing value and bearing 
value in the thinnest piece. The number of rivets in the raf- 
ter likewise depends upon the force it carries, unless the two 
rafters are supposed to abut and to transmit so much of the 
horizontal component as does not come through the inclined 
ties, a treatment not to be commended. The two angle-irons 
of the rafter, being in compression, should be connected at 
intervals by a rivet and filling piece or thimbje. The number 
of rivets through the rafter and connection plate at I. need only 
be enough to transmit the force from one diagonal to the raf- 
ter. Study the necessity for rivets, and do not add all the 
rivets in abutting pieces to obtain the number in a main 
member. 

Similarly, in IV., the first four or possibly five rivets on 
the left in the horizontal member balance the rivets in the 
inclined tie on the right; the six remaining rivets seen and 
three others unseen, on the left of the splice, balance the same 
number in the smaller angle. Note how, by an extension of 
the connecting plate and a short plate below, the main tie is 
neatly spliced and reduced in section. 

The rafter at III. has more rivets than at the upper end 
because the thrust is somewhat greater. The rivets in the tie 
at that connection will practically equal those at the other end 
of the same piece. The black holes at VI. indicate the rivets 
to be inserted at the time of erection, and these should, in good 
practice, exceed the number called for in joints riveted in the 
shop. They must carry the load and resist the moment of the 
horizontal component due to the wind pressure, which passes 
down the post IX. as shear. The post is subjected to bend- 
ing moment as well as compression, and hence has one dimen- 
sion much greater than the other. Bracing perpendicular to 
the plane of the truss is needed to resist wind pressure on the 
end of the structure. Columns and compression members, in 
structural work of any kind, if joined one to another, must be 
thoroughly stayed against lateral movement. 



2 66 STRUCTURAL MECHANICS. 

Pin-connected roof trusses resemble in their details 
the joints of the next section. 

273. Pin-Connected Bridge. — Ordinary details in a pin- 
jointed bridge truss of moderate span are shown in VII., 
VIII. and XVI. The position of the splice in the top chord 
is near the pin. The splice plate may be extended to rein- 
force the pinhole, if required. The ends of the chord pieces 
are machined plane and parallel, and only enough rivets are 
then used in the splice to insure the alignement. The pin is 
placed in the centre of gravity of the chord section, unless 
slightly changed from that position for the same reason as is 
given under § 130. -The connection plates are seen below, to 
keep the sides of the chord from spreading; the rest of the 
panel length is usually laced. Another chord section, em- 
ploying channels, is drawn at XI. 

The post has been discussed in Chapter IX. XII., XIII. 
and XIV. show other sections for posts. They offer facilities 
for the central support of floor-beams. Post flanges are 
sometimes turned out, sometimes in. The floor-beam, of 
plate girder type, is riveted at XVI. to the post through the 
holes shown. This attachment stiffens the trusses laterally 
and is much superior to hangers. Top and bottom lateral 
bracing, to convey the wind pressure to the abutments, is 
needed in the planes of the chords, and portal bracing at 
each end to throw the wind pressure from the top system into 
the end posts, which convey it to the abutments as shear, with 
the accompanying bending moments in those posts. 

The posts go inside of the top chord, as do the main 
diagonals or ties, which come next to the posts. The bottom 
chord bars are on the outside, one of those running towards 
the middle of the span being usually the farthest out. The 
bending moment on the pin was discussed in § 224. 

274. Riveted Bridges. — A riveted Warren girder or 
latticed truss is shown below. These details are not for con- 
secutive joints. The increase of chord section, when neces- 
sary, is indicated at XIX. If the truss is loaded on the top, 
interior diagonal bracing, drawn at XXL, must be used. 
When the truss is a lattice, the web members are connected 
at intersections to stiffen the compression members, as at 




xnr. 



L Xx.^ 




268 



STRUCTURAL MECHANICS. 



XX., or preferably as at V., or at XV., if the web is double. 
Horizontal lateral bracing must* not be overlooked. 

X. is one form of section of a solid bridge floor. Rec- 
tangular sections are also used. 

Examples. — i. Four timbers, 6 in. X 12 in., 15 ft. long, are in 
compression. If placed side by side, with space between for circu- 
lation of air, what will be the max. permissible distance between 
packing pieces, so that the timbers may be equally safe against 
flexure in either direction ? 

2. What pull can the bolt of IX., Plate II., safely resist, if 
the nut is i^ in. square, is 8 in, from end of stick, and q = 150 
lbs. ? If the compression under the nut ought not to exceed 1,800 
lbs., what can the bolt. carry without deducting its cross-section ? 

7,200 lbs. 4,050 lbs. 

3. hxi 8 in. by 8 in. timber of Southern pine is spliced as in 
XL, Plate II. Design the splice by § 169, 4th line from bottom 
■of table, and find what it will carry, neglecting the bolts. 

4. A cast beam, 8 ft. long, supported at ends, has to carry 
900 lbs. per ft. If /t =: 5,000 lbs.,/c = 13,500 lbs., and the 
section is I shaped, all parts ^ in. thick, what will be the breadth 
of top and bottom flanges to resist all the bending moment, if the 
effective depth at middle is 8 in.? If the flange sections are con- 
stant, what will be the elevation of the beam, and the depth at 
quarter span? 1.6 in., 4 32 in.; 6 in. 

5. A vessel is 200 ft. long. It carries 5 tons per ft. uniformly 
distributed, and a central load of 300 tons. Find M max. when 
at rest; when supported on a wave crest at bow and stem with each 
bearing 20 ft. long; and when supported amidships only with 
bearing 30 ft. long. 

6. The end of a beam 6 in. wide is built into a wall 18 in. 
The bending moment at the wall is 600,000 in. lbs. If the top of 
the beam bears for 9 in. with a uniformly varying pressure and 
the bottom the same, what is the max. unit compression on the 
bearing surface? 1,852 lbs. 

7. A plate girder for draw-span, 8 ft. 3 in. in outside depth, 
has a web of section, 96 X^ in., weight 120 lbs. per ft.; 4 flange 
angles, each 6x6x1 in., 39.2 lbs. per ft., I for one angle = 43. i, 
distance of centre of gravity from back of angle 1.96 in.; 15 in. 
channel on top and bottom, 60 lbs. per ft., i in. web, flanges 
turned in, I = 23.0, distance of centre of gravity from back, 
0.95 in.; and a 15 X j^ in. flange plate of 25 lbs. on each flange. 
Sketch the section, find I, and the weight per ft. Add 360 lbs. per 
ft. If the girder is 95 ft. long and fixed at one end only, what 
are /and v max.? 

I = 247,447; w — 447 lbs.;/= 8,740 lbs. V = 2.2 in. 



INDEX. 



PAGE 

Alteration of structure 161 

Arches, brick, concrete 110 

Ashlar masonry 36 

Axis, neutral 65 

Axes of direct stress in a beam 196 

of symmetry 89 

Bauschinger's endurance tests 160 

experiments 159 

laws 166 

Beams 44 

axes of direct stress , 196 

bending moment, see moment 47 

built wooden 262 

cement 109 

Clapeyron's formula 121 

compound 62 

continuous 121 

curved TO 

deflection 96 

distribution of internal stress 63 

shear 75 

flexure 96 

formulas 5i 

I beams 240 

impact 112 

inclined 69 

internal work Ill 

local loading 196 

lowering middle support 118 

moments 47 

reactions 44 

restrained 113 

resilience 110 

sandwich... 109 

shear 49 

strut 152 

tieand 132 

twisting 134 

. two spans 117 

uniform strength 72 

weight 74 

wooden 261 

Bearing plates 182 

Bending and torsion 83 

Bending moment, see Moment 47 

Blocks in compression 17, 138 

Boilers 217 

rivets 212 

Breaking strength 10 

Bricks 37 

Burnettizing 21 

Carbon 22 

in cast-iron 22 

in steel 25 

Castings, steel 32 

Cast-iron 22 

specifications 29 

Cement, beams of 109 

natural 40 

Portland 41 

specifications 42 

Change of length 6 

Channels, spacing of 04 

Clapeyron's formula 121 

Clay, brick 37 

pipe 23J^ 

Coefficient of elasticity 6. 199. 201 



PAGE 

Coefficients for continuous beams 124 

Columns 141 

direction of flexure 142 

eccentrically loaded 148 

Euler's formula 144 

experimental results 147 

failure by tension 147 

fixed or hinged ends 146 

formula 142, 144, 176 

Gordon's formula 176 

hinged ends.. .'. 145 

lacing 153 

multipliers of a 145 

pin ends 146 

Rankine's formula 145 

resistance. I4i . 

short 147 

slender 147 

straight line formula 149 

swelled 148 

tension ]47 

timber 173 

transverse force 152 

yield point 142 

Compression, blocks in 138 

curve 12 

ductile substances under 16 

fibrous substances under 17 

granular substances under 15 

load not central 138 

shear due to 3 

vitrious substances under 17 

Concrete . 41 

strength of — and steel 109 

Connecting rod 133 

Cooper's lines 204 

Corrugated iron 256 

Cross-section of equal strength 71 

Curvature of a beam 97 

Curve, compression 12 

stress-stretch 7 

Cylinders, thick 222 

thin 217 

Decay of wood 20 

Deflection of beams 96 

cantilever 98, 99 

supported at ends 100. 101, 102 

uniform strength 105 

Domes 234 

Drifting test for steel 33 

Ductile substances under compression.. 16- 

Dynamic theory, Fidler's 164 

Earth pressure 247 

Eccentric load 129, 138, 148 

Elasticity, Modulus of & 

shearing 7, 199 

Elastic limit 8 

alternating stresses 161 

conmiercial or common 9 

raised 10 

Ellipse of stress 190 

Elongation of steel, work of.. 7, 10, 11, 13, 28 

Envelopes 217 

Euler s formula 144 

Eyebars 135 

Fibrous substances under compression. 17 
Fidler's dynamic theory I6t 



270 



INDEX. 



PAGE 

Forces, external 1,2 

Gerber's Parabola 166 

Girders — see Beams 44 

Plate 240 

Granular substances under compres- 
sion 15 

Hooks 132 

Hoops 226 

I Beam 240 

Impact 170,175 

Inertia, Moment of. See Moment 85 

Internal stress 2, 183 

Iron, carbon in 22 

cast 22 

chilled 23 

column 179 

malleable 27 

modulus of elasticity 23 

specifications 29 

wrought 24 

case-hardened 27 

double refined .' 25 

hardening 26 

modulus of elasticity 25 

specifications . 29 

working stresses 174,179, 181 

Joint, lap , 209 

riveted 206 

strength 212 

Knots in timber 19 

Lacing bars 153 

Launhardt-Wyrauch formula 162 

Lime 38 

Load not central 129, 138, 148 

sudden application of 14, 170 

Loading, local on a beam 196 

Masonry . 35 

ashlar 36 

laid in winter 43 

Middle third ' 139 

Modulus of elasticity 6 

of volume 201 

resilience 110 

rupture 68 

shearing elasticity 7, 199 

Moment, bending 47, 52 

maximum, point of 52 

loads, moving 61 

pins 214 

wheel concentration 59 

of inertia 85 

axes of symmetry 89 

oblique axis 90 

polar 80 

spacing of channels 94 

thin sections 91 

resisting 66 

distribution of internal stress... 64 

limit of application 68 

oblique axis . . , 90 

tension equals compression 63 

torsional 80 

Mortar ' 39 

Neuiral axis 65 

movement of 70 

Nickel in steel 32 

Nuts 1.35 



Parabola, Gerber's 

Pedestals 


166 

182 


Permanent set 


8 


Phosphorous in iron and steel.. . 

Pier moment coefficients 

Pins . . 


. . 26, 30, 32 

125 

213 


allowable stress 

bearing . . . . 


181 

213 


distribution of shear in 


75 


shear 


213 


Piston, conical 


232 


Plaster 


40 



PAGE 

Plate girder 240 

Plates, bearing 182 

resistance of 253 

Polar moment of inertia 80 

Portland cement 41 

Posts — see Columns 138 

Preservation of wood 21 

Pressure of earth 247 

Pull and thrust 188 

Punching rivet holes 207 

test of steel 33 

Rankine's formula 145 

Reactions 44, 50 

Reinforcing plate 213 

Resilience 13 

of a beam 110 

Resistance of large blocks to crushing.. . 17 

of columns 141 

of a thin ring 235 

Resisting moment— see Moment 66 

Ring, resistance of 235 

Rivets, allowable stress 181 

arrangement 209 

bearing 206 

bending 207 

boiler 212 

diameter 207 

friction 207 

number 208 

plate girder 243 

resistance of plate 206 

shear 206 

size 211 

spacing 207 

specifications 211 

steel 212 

structural 211 

working stresses 181 

Riveted joints 206 

Rollers 182 

Safe working stresses 1.57 

Sandwich beams 109 

Screw threads 135 

vSecondary stresses 211 

Seefehlner's rule 166 

Segmental head 219 

Set, permanent 8 

Sewer pipe 238 

Shear 3, 49, 51, 186 

change of principal stress by 194 

sign 52 

compression and 3 

distribution in section of beam 75 

effect of two 198 

in beam 51 

maximum 53 

position of wheels for 60 

modulus . 199 

pins . .. .75, 213 

planes at right angles 184 

tension and 3 

variation of unit 77 

Shearing planes 191 

Shrinkage of timber 20 

Sign of shear 49 

stress 3 

Span 45 

Specifications for cast-iron 29 

cement 42 

masonry 35 

steel 30, 173 

timbpr 22 

Sphere, thick, hollow 227 

Spherical shell 219 

Splices 209 

in plate 134 

limber 258 

Springs 105,251 

Stand pipe 221 



INDEX. 



271 



Steel 25 

chemical specifications 32 

classification 28 

machinery 81 

manipulation 32 

nickel 32 

punching and drifting 33 

rivet 212 

shearing 34 

specifications 30, 173 

structural ... 33 

ultimate strength 33 

working stresses 176, 179, 181 

Stiffeners 2i4 

Stone 34 

pressure on 182 

Strength, cross section of equal 71 

ultimate or breaking 10 

Stresses, alternating 161, 179 

changed by shear 194 

combined 184 

compound 185 

conical piston 232 

conjugate 185 

ellipse of 190 

internal. 183 

oblique section 183 

on any plane 188 

principal 185, 187 

effect of two 198 

to find 191 

reduction of unit 169 

resolvable 200 

reversal 161, 179 

safe working 1.57 

secondary 211 

shearing 186 

sign of 3 

unit 4 

working, bearing 181 

compressive 179 

shearing 181 

tensile 176 

Stress-stretch diagram, 7 

Struts— see Columns ' 141 

Sulphur in steel 26, 32 



PAGE 

Sudden load 14 

Tanks 228 

Tension 129 

connections 134 

eccentric pull 129 

shear and 3 

torsion and 134 

Three moment theorem 121 

Tie.... 2, 129 

and beam 132 

Timber 19 

breaking stresses 22 

column formula 173 

decay 20 

framing 258 

shrinkage 20 

specifications 22 

strength 21 

working stresses 172 

Torsion 80 

angle of 82 

bending and 83 

cylinder 80 

square shaft . . . .< 81 

St. Venant's equations 82 

tension and 134 

Trees, growth of 19 

Tubes, collapsing of 220 

Twist of shaft 82 

Ultimate strength 10 

Uniforna strength, beams of 72 

Unit stresses 4 

Varying cross-section, effect of 12 

Vitrious substances under compression. 17 

Volume, change of 197 

coefficient of elasticity of 12 

Wall, retaining 247 

Water pipes, cast iron for 29 

Web of plate girder 241 

Wohler's experiments 157 

laws 158 

Wood, decay and preservation 20, 21 

Work of elongation 7, 10, 13, 28 

internal, in a beam Ill 

Wrought iron — see Iron. 24 

Yield point 9 



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